978-960-7182-07-4

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Συνήθεις Διαφορικές Εξισώσεις 2η Αναθεωρημένη Έκδοση Νικόλαος Σταυρακάκης ISBN : 978-960-7182-07-4 Απαντήσεις Ασκήσεων - Προβλημάτων Περιεχόμενα Κεφάλαιο 1………………………………….. σελ. 2 Κεφάλαιο 2………………………………….. σελ. 3 Κεφάλαιο 3………………………………….. σελ. 13 Κεφάλαιο 4………………………………….. σελ. 14 Κεφάλαιο 5………………………………….. σελ. 19 Κεφάλαιο 6………………………………….. σελ. 31 Κεφάλαιο 7………………………………….. σελ. 38 Κεφάλαιο 8………………………………….. σελ. 48 Κεφάλαιο 9………………………………….. σελ. 51 Κεφάλαιο 10………………………….…….. σελ. 59

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Page 1: 978-960-7182-07-4

Συνήθεις Διαφορικές Εξισώσεις 2η Αναθεωρημένη Έκδοση

Νικόλαος Σταυρακάκης

ISBN 978-960-7182-07-4

Απαντήσεις Ασκήσεων - Προβλημάτων

Περιεχόμενα

Κεφάλαιο 1helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 2

Κεφάλαιο 2helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 3

Κεφάλαιο 3helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 13

Κεφάλαιο 4helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 14

Κεφάλαιο 5helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 19

Κεφάλαιο 6helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 31

Κεφάλαιο 7helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 38

Κεφάλαιο 8helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 48

Κεφάλαιο 9helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 51

Κεφάλαιο 10helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip σελ 59

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 1

11

22 6y = 2 e3x ndash 3 x2 + c 23 y = lnx + c

24 2y = ex2+ c 25

y = x sinndash1 x ndash x2 + 1 + c

26 y = x ex ndash ex + 3 27 y = sin2 x + 1

28 y = xln x ndash x 29 y = x2 + x + 3

30 3y = 2 x32 ndash 16 31 y = 10 tanndash 1 x

32 y = sinndash 1 x 33 y = c + ln x x ndash 2 ndash1

34 y = c1 + c2 x + c3 x2 + c4 x3 + 16 sin h x 2

35 y = c1 + c2 x + endashx ndash cos 3x 36 y = x2

37 y = ex 2 38 y = x

13

1 y = c xndash3 2

12

ln 2 y2 + x y + x2 + 17

arctan4y + xx 7

= c

3 1 ndash y2 ndash ln

1 + 1 ndash y2

y = x + c

4 4 y32 = ndash 3x + c

5 2 y2 = 2 ln x ndash x 2 + c 6 x2 ndash 2xy ndash y2 = c

7 c ey = y ndash x + 2

8

ex sin y = c

9

x = c cos2 y 10

2 ln y = x2 + y2 + c

11 2 x2 + 3 y2 = c

12 y2 ln y + x2 = c y2

13 2 y2 = 2 ln x + x2 + c

14 y = 14

ndash 16

x2 + c xndash 4

15 2 ln cos h y + x2 = c 16 y53 = x53 + c

ΚΕΦΑΛΑΙΟ 2

21

1 ναι 2 ναι

3 ναι 4 ναι

5 όχι 6 όχι

7 y t = tan t ndash 12 t2 + c 8 y t = arcsin c sin t

9 y2 + c y2 sin x + 1 = 0 10 y = ln cos c ndash x + k

11 y = c 1 ndash c cos x ndash1 12 ln 1 + y = x + x22 + c

13 1

3 y3ndash

2y =

1x + ln x + c 14 y = c x ndash3 2 x ndash1 ndash1

15 y = c + c1 x ndash a x ndash b1

a ndash b

16 γενική a2

ndash x2

= c cos2

y με x2 ne a

2 y ne π 2 + nπ

ιδιάζουσα y = π 2 + nπ

17 γενική ex + 1

+ ln csc y ndash cot y + cos y = c y ne n π ιδιάζουσα y = n π

18 γενική ln x y + 1 + y2 = c xne 1 x ne 0 ιδιάζουσες x=1 x=0

19 γενική sin y = x ndash 12

e2 x + c

x ne 1 y ne n π ιδιάζουσα y = nπ

20 γενική x2

+ 2x = endashy 2

+ c x ne ndash1 ιδιάζουσα x=ndash1

21 y t = 9 + 2 ln1+t2

5

1 21 2

ndashinfinlt t lt infin

22 y t = 1 ndash 4 + 2t + 2 t2 + t3 1 21 2 ndash2 lt t lt infin

23 a = b y t =

a2 kt1 + a kt

ndash1a k

lt t lt infin

a ne b y t =

a b 1 ndash ek b ndasha t

a ndash bek b ndash a t

ln abk b ndash a

lt t lt infin

24 y t = 2 et 2 et ndash1 tisinΙR 25 y = a 1 + 1 ndash x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

26 y equiv 0 27 cos2x cos2y = ndash 1

28 3 y2 = 1 + 2 e3 x2 ndash 12

29 y = 1 ndash x3 + 2x2 + 2x + 4 xisin ndash2 +infin 30 ln y + y2 = sin x + 1 y gt 0

32

y x =1 ndash x

2 2

για x le ndash1

2 1 ndashx2 2

για ndash 1 le x le 1

0 για 1 le x

34 y x =c1 x3 x ne 0

c2 x3 x ne 0 Για c1 = c2 = 1 η y(x)=x3 xisinIR ειδική λύση

36 a y = 1 ndash 2x ndash ln |c ndash x| b y = 1 + x + 1 + 2 c e x 1 ndash c endash3xndash3

c 2y ndash2x + sin2 (x + y) = c d 4 (y ndash 2x + 3) = (x + c)2

38 d y3 ndash 4 y ndash x3 = ndash 1 x3 ndash 1 lt16

3 3ή ndash 128 lt x lt 16

42 a y =ag

x +bg ndash ad

g2ln gx + d + c gne 0 gx + dne 0

b

x =ga y +

ad ndash bg

a2

ln ay + b + c a ne 0 ay + b ne 0

43 y = ndash x2y

x2+ 2y2 = krArr 44 y = rArr4yx y = k x4

45 x x ndashk 2 + y2 ndash k 2 + 1 = 0 k2 gt 1 46 y = rArrxy x2 ndash y2 = k

47 x ndash y = k x + y 3 48 x2 ndash 2ky=k2 kgt0

22

1 ex + (x 22) + (y 22) 2 όχι πλήρης

3 x3y ndash xy3 + (y 22) = c 4 t2 + z2 = c

5 ey ndash2 x2y ndash log x = c 6 όχι πλήρης

7 όχι πλήρης 8 x2 y ndash tanx + y2 = c

9 y = c ndash x 2 + xex ndash1 10 y = c ndash 3x x2 ndash 1ndash1

11 όχι πλήρης 12 sin x cosy + x2 ndash y2 = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 5

13 r = c ndash eθ sec θ 14 y = c + et t ndash 1 1 + et ndash1

15 x2 + x y2 ndash sin x + y ndash e y = c 16 όχι πλήρης

17 cos xyndash1 = c ή xyndash 1= c 18 όχι πλήρης

19 ln1 + xy

1 ndash xyndash 2x = c 20 x ey2

+ csc y cotx = c

21 x2 + y2 = c 22 x3

1 + ln y ndashy2

= c

23 3 x2y + y3 = 32 24 x2 + 2 xy ndash y2 = ndash16

25 x sin y + y2 = π2 26 x2 e2y + 2y = 4

27 y = x + 28 3x2 2 x lt 283 28 y = x ndash 24x3 + x2 ndash 8x ndash 16 1 21 2 4

29 xex y3 + ex ndash 6y3 = ndash5 30 2 cosx cosy = 1

31 exy2+ x4 ndash y3 = 2 32 y = e ndash x ex ndash1 ndash1 33 y = ndash2 tet + 2

ndash1

34 a Ν(x y) = xcos xy + g y b N x y = x exy ndash x 4 + g y όπου g αυθαίρετη

συνάρτηση του y

36 a k = 3 x2 y2 + 2x3y = c b k = 1 x2 + e2 xy = c

37 2x = y3 + cy 38 xy ndash x ndash 2y = c

39 x3 y + ey = c 40 x + arc tan y xndash1 = c

23

1 μ = yndash 4

x2y

ndash3ndashy

ndash1= c y equiv 0 2 μ = x

ndash1 xy ndash ln x ndash 2

ndash1y

2= c x equiv 0

3 μ =1

x3 y3 ndash

1

2x2 y2+

3

2y2 = c xequiv0 y equiv 0

4 μ t = et y t = ndash et plusmn e2t + 2 cendash t 1 21 2

5 μ t = endash at cosy a = y t = arcsin c ndasht et

6 μ = ty lny

t+ ty = c t equiv 0 y equiv 0

7 μ=t

ndash2y

ndash3

t2

y2

+ 1t = c t equiv 0 y equiv 0

6 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

8 μ=ndash tndash1

yndash2

2ln ty + ty = c t equiv 0 y equiv 0

9 μ = xy x3y + 3x2 + y3 = c

10 μ = x2

ndash y2 ndash1

x ndash y + ln x + y ndash ln x ndash y = c y = x y = ndashx

11 μ = endash y 2 xex ndash y + y2 = c 12 μ = x y ndash 4 34 3 y ndash x + 1 3 = c xy xequiv 0 y equiv 0

13 μ = xy x + y + 1 ndash 1 x + y + 1 3 = c xy xequiv 0 y equiv 0

14 μ = x 6x2 y2 + 8x3y + 3x4 = c x0

15 μ = x2

+ y2 1

y + arc tanyx = c y equiv 0

16 μ = xa yb a = b = x7 y2 ndash x3 y9 = c x 0 y 0

17 μ = eax

eby

a = b = x2

ndash 2xy endash x + 2y

= c xequiv0

18 μ = xa

ebx

a = b = x

2

2ye

6x+

e6x

96x ndash1 = c xequiv0

19 μ = xa

yb

a = b = x6

y4

+ x5

y7

= c xequiv0 yequiv0

20 μ = xa yb a = b = x3 y4 = 1296

21 μ = yaexb

a = b = y3ex2

= 125 e9

22 μ = xa

yb

ex

g

a = b = g = yx2

ex

2

= 12 e4

23 μ = x x2 y3 ndash 2 x2 = ndash 1143

24 μ = exbb = xy ex2

= ndash 2e 25 μ = σταθερά 4x2 + 3y2 = 247

26 μ = xa eb a = b = 3 x23 yex = ndash 21e

24

1 y = endashx

arc tanndash1

ex

+ cendashx

2 y = 1 + x2 ndash1

ln sin x + c 1 + x2 ndash1

3 y = x2 endashx + x2 ndash 2x + 2 + cendashx 4 y = x2 csc x + c csc x

5 xy sin x = sin x ndash xcosx + c 6 y = x3

+ c ln x3

7 y =cx +

3cos2x

4x+

3

2sin 2x 8 y = arc tan x + c 1 + x

2 ndash2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 7

9 y = 1 ndash

12

endash2x ndashinfin lt x lt infin

10 y =

qp + y0 ndash

qp endashp x ndash x

0 xisinΙR

11 y = x2

ndash 52x

0 lt x lt infin

12 y = x + x02 y0 ndash x0 x2 0 lt x lt infin αν x0 gt 0 και ndashinfin lt x lt 0 αν x0 lt 0

13 y = x + ln xx0 + x0 y0 ndash x0

2 x οltxltinfin αν x0gt0 και ndashltxlt0 αν x0lt0

14 y = a + ln 1 + x endash x

ndash1 lt x lt infin

15 y = 14

x2 ndash 13

x + 12

+ 112

xndash2 lt x lt infin 0

16 y = tanx + a sec x ndashπ22 lt x lt π22

17 y(x)0 όταν λlt0 y(x) y0 όταν λ=0 και y(x) +infinόταν λgt0

19 y(x) = x2 όταν x0 και y(x) = x2ndash2x3 όταν xge0

20 y =

x ndash 13

6+ c1 ln x ndash 1 + c2 21 y x =

ex x lt 02ex ndashx ndash1 x ge 0

22 y x =endash x x + 1 0 le x lt 2

2endash x + endash2 23 y x =

12

1 ndash endash 2x 0 le x le 1

12

e2 ndash1 endash2 x 1 lt x

24 y x =endash2x 0 le x le 1

endash x + 1 1 lt x 27 ναι αν y0 28 ναι 29 ναι

30 όχι 31 ναι 32 ναι 33 ναι 34 ναι 35 y = x2 ln x + cx2

36 y = sin x + c cosx 37 y =12

ndash1x +

c

x2

38 y =x2 x ndash 1

2+ c 1 ndash x 39 y = endashx +

cendashx

x

40 3y4 x8 ndash 2x6 = c 41 x2 yndash2 + 2 ex = c

42 ndash 3 x2 yndash 2 32 3 ndash 4x = c 43

u = xyndash 2 y = 23 + cx2

44 y =plusmn k l + cke ndash2kxndash 1 2

45 y = plusmn σ 12x 2 σ s ds + c

xndash12

όπου σ x = exp 2Γ sin x + 2Tx

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 2: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 1

11

22 6y = 2 e3x ndash 3 x2 + c 23 y = lnx + c

24 2y = ex2+ c 25

y = x sinndash1 x ndash x2 + 1 + c

26 y = x ex ndash ex + 3 27 y = sin2 x + 1

28 y = xln x ndash x 29 y = x2 + x + 3

30 3y = 2 x32 ndash 16 31 y = 10 tanndash 1 x

32 y = sinndash 1 x 33 y = c + ln x x ndash 2 ndash1

34 y = c1 + c2 x + c3 x2 + c4 x3 + 16 sin h x 2

35 y = c1 + c2 x + endashx ndash cos 3x 36 y = x2

37 y = ex 2 38 y = x

13

1 y = c xndash3 2

12

ln 2 y2 + x y + x2 + 17

arctan4y + xx 7

= c

3 1 ndash y2 ndash ln

1 + 1 ndash y2

y = x + c

4 4 y32 = ndash 3x + c

5 2 y2 = 2 ln x ndash x 2 + c 6 x2 ndash 2xy ndash y2 = c

7 c ey = y ndash x + 2

8

ex sin y = c

9

x = c cos2 y 10

2 ln y = x2 + y2 + c

11 2 x2 + 3 y2 = c

12 y2 ln y + x2 = c y2

13 2 y2 = 2 ln x + x2 + c

14 y = 14

ndash 16

x2 + c xndash 4

15 2 ln cos h y + x2 = c 16 y53 = x53 + c

ΚΕΦΑΛΑΙΟ 2

21

1 ναι 2 ναι

3 ναι 4 ναι

5 όχι 6 όχι

7 y t = tan t ndash 12 t2 + c 8 y t = arcsin c sin t

9 y2 + c y2 sin x + 1 = 0 10 y = ln cos c ndash x + k

11 y = c 1 ndash c cos x ndash1 12 ln 1 + y = x + x22 + c

13 1

3 y3ndash

2y =

1x + ln x + c 14 y = c x ndash3 2 x ndash1 ndash1

15 y = c + c1 x ndash a x ndash b1

a ndash b

16 γενική a2

ndash x2

= c cos2

y με x2 ne a

2 y ne π 2 + nπ

ιδιάζουσα y = π 2 + nπ

17 γενική ex + 1

+ ln csc y ndash cot y + cos y = c y ne n π ιδιάζουσα y = n π

18 γενική ln x y + 1 + y2 = c xne 1 x ne 0 ιδιάζουσες x=1 x=0

19 γενική sin y = x ndash 12

e2 x + c

x ne 1 y ne n π ιδιάζουσα y = nπ

20 γενική x2

+ 2x = endashy 2

+ c x ne ndash1 ιδιάζουσα x=ndash1

21 y t = 9 + 2 ln1+t2

5

1 21 2

ndashinfinlt t lt infin

22 y t = 1 ndash 4 + 2t + 2 t2 + t3 1 21 2 ndash2 lt t lt infin

23 a = b y t =

a2 kt1 + a kt

ndash1a k

lt t lt infin

a ne b y t =

a b 1 ndash ek b ndasha t

a ndash bek b ndash a t

ln abk b ndash a

lt t lt infin

24 y t = 2 et 2 et ndash1 tisinΙR 25 y = a 1 + 1 ndash x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

26 y equiv 0 27 cos2x cos2y = ndash 1

28 3 y2 = 1 + 2 e3 x2 ndash 12

29 y = 1 ndash x3 + 2x2 + 2x + 4 xisin ndash2 +infin 30 ln y + y2 = sin x + 1 y gt 0

32

y x =1 ndash x

2 2

για x le ndash1

2 1 ndashx2 2

για ndash 1 le x le 1

0 για 1 le x

34 y x =c1 x3 x ne 0

c2 x3 x ne 0 Για c1 = c2 = 1 η y(x)=x3 xisinIR ειδική λύση

36 a y = 1 ndash 2x ndash ln |c ndash x| b y = 1 + x + 1 + 2 c e x 1 ndash c endash3xndash3

c 2y ndash2x + sin2 (x + y) = c d 4 (y ndash 2x + 3) = (x + c)2

38 d y3 ndash 4 y ndash x3 = ndash 1 x3 ndash 1 lt16

3 3ή ndash 128 lt x lt 16

42 a y =ag

x +bg ndash ad

g2ln gx + d + c gne 0 gx + dne 0

b

x =ga y +

ad ndash bg

a2

ln ay + b + c a ne 0 ay + b ne 0

43 y = ndash x2y

x2+ 2y2 = krArr 44 y = rArr4yx y = k x4

45 x x ndashk 2 + y2 ndash k 2 + 1 = 0 k2 gt 1 46 y = rArrxy x2 ndash y2 = k

47 x ndash y = k x + y 3 48 x2 ndash 2ky=k2 kgt0

22

1 ex + (x 22) + (y 22) 2 όχι πλήρης

3 x3y ndash xy3 + (y 22) = c 4 t2 + z2 = c

5 ey ndash2 x2y ndash log x = c 6 όχι πλήρης

7 όχι πλήρης 8 x2 y ndash tanx + y2 = c

9 y = c ndash x 2 + xex ndash1 10 y = c ndash 3x x2 ndash 1ndash1

11 όχι πλήρης 12 sin x cosy + x2 ndash y2 = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 5

13 r = c ndash eθ sec θ 14 y = c + et t ndash 1 1 + et ndash1

15 x2 + x y2 ndash sin x + y ndash e y = c 16 όχι πλήρης

17 cos xyndash1 = c ή xyndash 1= c 18 όχι πλήρης

19 ln1 + xy

1 ndash xyndash 2x = c 20 x ey2

+ csc y cotx = c

21 x2 + y2 = c 22 x3

1 + ln y ndashy2

= c

23 3 x2y + y3 = 32 24 x2 + 2 xy ndash y2 = ndash16

25 x sin y + y2 = π2 26 x2 e2y + 2y = 4

27 y = x + 28 3x2 2 x lt 283 28 y = x ndash 24x3 + x2 ndash 8x ndash 16 1 21 2 4

29 xex y3 + ex ndash 6y3 = ndash5 30 2 cosx cosy = 1

31 exy2+ x4 ndash y3 = 2 32 y = e ndash x ex ndash1 ndash1 33 y = ndash2 tet + 2

ndash1

34 a Ν(x y) = xcos xy + g y b N x y = x exy ndash x 4 + g y όπου g αυθαίρετη

συνάρτηση του y

36 a k = 3 x2 y2 + 2x3y = c b k = 1 x2 + e2 xy = c

37 2x = y3 + cy 38 xy ndash x ndash 2y = c

39 x3 y + ey = c 40 x + arc tan y xndash1 = c

23

1 μ = yndash 4

x2y

ndash3ndashy

ndash1= c y equiv 0 2 μ = x

ndash1 xy ndash ln x ndash 2

ndash1y

2= c x equiv 0

3 μ =1

x3 y3 ndash

1

2x2 y2+

3

2y2 = c xequiv0 y equiv 0

4 μ t = et y t = ndash et plusmn e2t + 2 cendash t 1 21 2

5 μ t = endash at cosy a = y t = arcsin c ndasht et

6 μ = ty lny

t+ ty = c t equiv 0 y equiv 0

7 μ=t

ndash2y

ndash3

t2

y2

+ 1t = c t equiv 0 y equiv 0

6 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

8 μ=ndash tndash1

yndash2

2ln ty + ty = c t equiv 0 y equiv 0

9 μ = xy x3y + 3x2 + y3 = c

10 μ = x2

ndash y2 ndash1

x ndash y + ln x + y ndash ln x ndash y = c y = x y = ndashx

11 μ = endash y 2 xex ndash y + y2 = c 12 μ = x y ndash 4 34 3 y ndash x + 1 3 = c xy xequiv 0 y equiv 0

13 μ = xy x + y + 1 ndash 1 x + y + 1 3 = c xy xequiv 0 y equiv 0

14 μ = x 6x2 y2 + 8x3y + 3x4 = c x0

15 μ = x2

+ y2 1

y + arc tanyx = c y equiv 0

16 μ = xa yb a = b = x7 y2 ndash x3 y9 = c x 0 y 0

17 μ = eax

eby

a = b = x2

ndash 2xy endash x + 2y

= c xequiv0

18 μ = xa

ebx

a = b = x

2

2ye

6x+

e6x

96x ndash1 = c xequiv0

19 μ = xa

yb

a = b = x6

y4

+ x5

y7

= c xequiv0 yequiv0

20 μ = xa yb a = b = x3 y4 = 1296

21 μ = yaexb

a = b = y3ex2

= 125 e9

22 μ = xa

yb

ex

g

a = b = g = yx2

ex

2

= 12 e4

23 μ = x x2 y3 ndash 2 x2 = ndash 1143

24 μ = exbb = xy ex2

= ndash 2e 25 μ = σταθερά 4x2 + 3y2 = 247

26 μ = xa eb a = b = 3 x23 yex = ndash 21e

24

1 y = endashx

arc tanndash1

ex

+ cendashx

2 y = 1 + x2 ndash1

ln sin x + c 1 + x2 ndash1

3 y = x2 endashx + x2 ndash 2x + 2 + cendashx 4 y = x2 csc x + c csc x

5 xy sin x = sin x ndash xcosx + c 6 y = x3

+ c ln x3

7 y =cx +

3cos2x

4x+

3

2sin 2x 8 y = arc tan x + c 1 + x

2 ndash2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 7

9 y = 1 ndash

12

endash2x ndashinfin lt x lt infin

10 y =

qp + y0 ndash

qp endashp x ndash x

0 xisinΙR

11 y = x2

ndash 52x

0 lt x lt infin

12 y = x + x02 y0 ndash x0 x2 0 lt x lt infin αν x0 gt 0 και ndashinfin lt x lt 0 αν x0 lt 0

13 y = x + ln xx0 + x0 y0 ndash x0

2 x οltxltinfin αν x0gt0 και ndashltxlt0 αν x0lt0

14 y = a + ln 1 + x endash x

ndash1 lt x lt infin

15 y = 14

x2 ndash 13

x + 12

+ 112

xndash2 lt x lt infin 0

16 y = tanx + a sec x ndashπ22 lt x lt π22

17 y(x)0 όταν λlt0 y(x) y0 όταν λ=0 και y(x) +infinόταν λgt0

19 y(x) = x2 όταν x0 και y(x) = x2ndash2x3 όταν xge0

20 y =

x ndash 13

6+ c1 ln x ndash 1 + c2 21 y x =

ex x lt 02ex ndashx ndash1 x ge 0

22 y x =endash x x + 1 0 le x lt 2

2endash x + endash2 23 y x =

12

1 ndash endash 2x 0 le x le 1

12

e2 ndash1 endash2 x 1 lt x

24 y x =endash2x 0 le x le 1

endash x + 1 1 lt x 27 ναι αν y0 28 ναι 29 ναι

30 όχι 31 ναι 32 ναι 33 ναι 34 ναι 35 y = x2 ln x + cx2

36 y = sin x + c cosx 37 y =12

ndash1x +

c

x2

38 y =x2 x ndash 1

2+ c 1 ndash x 39 y = endashx +

cendashx

x

40 3y4 x8 ndash 2x6 = c 41 x2 yndash2 + 2 ex = c

42 ndash 3 x2 yndash 2 32 3 ndash 4x = c 43

u = xyndash 2 y = 23 + cx2

44 y =plusmn k l + cke ndash2kxndash 1 2

45 y = plusmn σ 12x 2 σ s ds + c

xndash12

όπου σ x = exp 2Γ sin x + 2Tx

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
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  • Κεφάλαιο 10
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Page 3: 978-960-7182-07-4

ΚΕΦΑΛΑΙΟ 2

21

1 ναι 2 ναι

3 ναι 4 ναι

5 όχι 6 όχι

7 y t = tan t ndash 12 t2 + c 8 y t = arcsin c sin t

9 y2 + c y2 sin x + 1 = 0 10 y = ln cos c ndash x + k

11 y = c 1 ndash c cos x ndash1 12 ln 1 + y = x + x22 + c

13 1

3 y3ndash

2y =

1x + ln x + c 14 y = c x ndash3 2 x ndash1 ndash1

15 y = c + c1 x ndash a x ndash b1

a ndash b

16 γενική a2

ndash x2

= c cos2

y με x2 ne a

2 y ne π 2 + nπ

ιδιάζουσα y = π 2 + nπ

17 γενική ex + 1

+ ln csc y ndash cot y + cos y = c y ne n π ιδιάζουσα y = n π

18 γενική ln x y + 1 + y2 = c xne 1 x ne 0 ιδιάζουσες x=1 x=0

19 γενική sin y = x ndash 12

e2 x + c

x ne 1 y ne n π ιδιάζουσα y = nπ

20 γενική x2

+ 2x = endashy 2

+ c x ne ndash1 ιδιάζουσα x=ndash1

21 y t = 9 + 2 ln1+t2

5

1 21 2

ndashinfinlt t lt infin

22 y t = 1 ndash 4 + 2t + 2 t2 + t3 1 21 2 ndash2 lt t lt infin

23 a = b y t =

a2 kt1 + a kt

ndash1a k

lt t lt infin

a ne b y t =

a b 1 ndash ek b ndasha t

a ndash bek b ndash a t

ln abk b ndash a

lt t lt infin

24 y t = 2 et 2 et ndash1 tisinΙR 25 y = a 1 + 1 ndash x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

26 y equiv 0 27 cos2x cos2y = ndash 1

28 3 y2 = 1 + 2 e3 x2 ndash 12

29 y = 1 ndash x3 + 2x2 + 2x + 4 xisin ndash2 +infin 30 ln y + y2 = sin x + 1 y gt 0

32

y x =1 ndash x

2 2

για x le ndash1

2 1 ndashx2 2

για ndash 1 le x le 1

0 για 1 le x

34 y x =c1 x3 x ne 0

c2 x3 x ne 0 Για c1 = c2 = 1 η y(x)=x3 xisinIR ειδική λύση

36 a y = 1 ndash 2x ndash ln |c ndash x| b y = 1 + x + 1 + 2 c e x 1 ndash c endash3xndash3

c 2y ndash2x + sin2 (x + y) = c d 4 (y ndash 2x + 3) = (x + c)2

38 d y3 ndash 4 y ndash x3 = ndash 1 x3 ndash 1 lt16

3 3ή ndash 128 lt x lt 16

42 a y =ag

x +bg ndash ad

g2ln gx + d + c gne 0 gx + dne 0

b

x =ga y +

ad ndash bg

a2

ln ay + b + c a ne 0 ay + b ne 0

43 y = ndash x2y

x2+ 2y2 = krArr 44 y = rArr4yx y = k x4

45 x x ndashk 2 + y2 ndash k 2 + 1 = 0 k2 gt 1 46 y = rArrxy x2 ndash y2 = k

47 x ndash y = k x + y 3 48 x2 ndash 2ky=k2 kgt0

22

1 ex + (x 22) + (y 22) 2 όχι πλήρης

3 x3y ndash xy3 + (y 22) = c 4 t2 + z2 = c

5 ey ndash2 x2y ndash log x = c 6 όχι πλήρης

7 όχι πλήρης 8 x2 y ndash tanx + y2 = c

9 y = c ndash x 2 + xex ndash1 10 y = c ndash 3x x2 ndash 1ndash1

11 όχι πλήρης 12 sin x cosy + x2 ndash y2 = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 5

13 r = c ndash eθ sec θ 14 y = c + et t ndash 1 1 + et ndash1

15 x2 + x y2 ndash sin x + y ndash e y = c 16 όχι πλήρης

17 cos xyndash1 = c ή xyndash 1= c 18 όχι πλήρης

19 ln1 + xy

1 ndash xyndash 2x = c 20 x ey2

+ csc y cotx = c

21 x2 + y2 = c 22 x3

1 + ln y ndashy2

= c

23 3 x2y + y3 = 32 24 x2 + 2 xy ndash y2 = ndash16

25 x sin y + y2 = π2 26 x2 e2y + 2y = 4

27 y = x + 28 3x2 2 x lt 283 28 y = x ndash 24x3 + x2 ndash 8x ndash 16 1 21 2 4

29 xex y3 + ex ndash 6y3 = ndash5 30 2 cosx cosy = 1

31 exy2+ x4 ndash y3 = 2 32 y = e ndash x ex ndash1 ndash1 33 y = ndash2 tet + 2

ndash1

34 a Ν(x y) = xcos xy + g y b N x y = x exy ndash x 4 + g y όπου g αυθαίρετη

συνάρτηση του y

36 a k = 3 x2 y2 + 2x3y = c b k = 1 x2 + e2 xy = c

37 2x = y3 + cy 38 xy ndash x ndash 2y = c

39 x3 y + ey = c 40 x + arc tan y xndash1 = c

23

1 μ = yndash 4

x2y

ndash3ndashy

ndash1= c y equiv 0 2 μ = x

ndash1 xy ndash ln x ndash 2

ndash1y

2= c x equiv 0

3 μ =1

x3 y3 ndash

1

2x2 y2+

3

2y2 = c xequiv0 y equiv 0

4 μ t = et y t = ndash et plusmn e2t + 2 cendash t 1 21 2

5 μ t = endash at cosy a = y t = arcsin c ndasht et

6 μ = ty lny

t+ ty = c t equiv 0 y equiv 0

7 μ=t

ndash2y

ndash3

t2

y2

+ 1t = c t equiv 0 y equiv 0

6 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

8 μ=ndash tndash1

yndash2

2ln ty + ty = c t equiv 0 y equiv 0

9 μ = xy x3y + 3x2 + y3 = c

10 μ = x2

ndash y2 ndash1

x ndash y + ln x + y ndash ln x ndash y = c y = x y = ndashx

11 μ = endash y 2 xex ndash y + y2 = c 12 μ = x y ndash 4 34 3 y ndash x + 1 3 = c xy xequiv 0 y equiv 0

13 μ = xy x + y + 1 ndash 1 x + y + 1 3 = c xy xequiv 0 y equiv 0

14 μ = x 6x2 y2 + 8x3y + 3x4 = c x0

15 μ = x2

+ y2 1

y + arc tanyx = c y equiv 0

16 μ = xa yb a = b = x7 y2 ndash x3 y9 = c x 0 y 0

17 μ = eax

eby

a = b = x2

ndash 2xy endash x + 2y

= c xequiv0

18 μ = xa

ebx

a = b = x

2

2ye

6x+

e6x

96x ndash1 = c xequiv0

19 μ = xa

yb

a = b = x6

y4

+ x5

y7

= c xequiv0 yequiv0

20 μ = xa yb a = b = x3 y4 = 1296

21 μ = yaexb

a = b = y3ex2

= 125 e9

22 μ = xa

yb

ex

g

a = b = g = yx2

ex

2

= 12 e4

23 μ = x x2 y3 ndash 2 x2 = ndash 1143

24 μ = exbb = xy ex2

= ndash 2e 25 μ = σταθερά 4x2 + 3y2 = 247

26 μ = xa eb a = b = 3 x23 yex = ndash 21e

24

1 y = endashx

arc tanndash1

ex

+ cendashx

2 y = 1 + x2 ndash1

ln sin x + c 1 + x2 ndash1

3 y = x2 endashx + x2 ndash 2x + 2 + cendashx 4 y = x2 csc x + c csc x

5 xy sin x = sin x ndash xcosx + c 6 y = x3

+ c ln x3

7 y =cx +

3cos2x

4x+

3

2sin 2x 8 y = arc tan x + c 1 + x

2 ndash2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 7

9 y = 1 ndash

12

endash2x ndashinfin lt x lt infin

10 y =

qp + y0 ndash

qp endashp x ndash x

0 xisinΙR

11 y = x2

ndash 52x

0 lt x lt infin

12 y = x + x02 y0 ndash x0 x2 0 lt x lt infin αν x0 gt 0 και ndashinfin lt x lt 0 αν x0 lt 0

13 y = x + ln xx0 + x0 y0 ndash x0

2 x οltxltinfin αν x0gt0 και ndashltxlt0 αν x0lt0

14 y = a + ln 1 + x endash x

ndash1 lt x lt infin

15 y = 14

x2 ndash 13

x + 12

+ 112

xndash2 lt x lt infin 0

16 y = tanx + a sec x ndashπ22 lt x lt π22

17 y(x)0 όταν λlt0 y(x) y0 όταν λ=0 και y(x) +infinόταν λgt0

19 y(x) = x2 όταν x0 και y(x) = x2ndash2x3 όταν xge0

20 y =

x ndash 13

6+ c1 ln x ndash 1 + c2 21 y x =

ex x lt 02ex ndashx ndash1 x ge 0

22 y x =endash x x + 1 0 le x lt 2

2endash x + endash2 23 y x =

12

1 ndash endash 2x 0 le x le 1

12

e2 ndash1 endash2 x 1 lt x

24 y x =endash2x 0 le x le 1

endash x + 1 1 lt x 27 ναι αν y0 28 ναι 29 ναι

30 όχι 31 ναι 32 ναι 33 ναι 34 ναι 35 y = x2 ln x + cx2

36 y = sin x + c cosx 37 y =12

ndash1x +

c

x2

38 y =x2 x ndash 1

2+ c 1 ndash x 39 y = endashx +

cendashx

x

40 3y4 x8 ndash 2x6 = c 41 x2 yndash2 + 2 ex = c

42 ndash 3 x2 yndash 2 32 3 ndash 4x = c 43

u = xyndash 2 y = 23 + cx2

44 y =plusmn k l + cke ndash2kxndash 1 2

45 y = plusmn σ 12x 2 σ s ds + c

xndash12

όπου σ x = exp 2Γ sin x + 2Tx

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
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Page 4: 978-960-7182-07-4

4 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

26 y equiv 0 27 cos2x cos2y = ndash 1

28 3 y2 = 1 + 2 e3 x2 ndash 12

29 y = 1 ndash x3 + 2x2 + 2x + 4 xisin ndash2 +infin 30 ln y + y2 = sin x + 1 y gt 0

32

y x =1 ndash x

2 2

για x le ndash1

2 1 ndashx2 2

για ndash 1 le x le 1

0 για 1 le x

34 y x =c1 x3 x ne 0

c2 x3 x ne 0 Για c1 = c2 = 1 η y(x)=x3 xisinIR ειδική λύση

36 a y = 1 ndash 2x ndash ln |c ndash x| b y = 1 + x + 1 + 2 c e x 1 ndash c endash3xndash3

c 2y ndash2x + sin2 (x + y) = c d 4 (y ndash 2x + 3) = (x + c)2

38 d y3 ndash 4 y ndash x3 = ndash 1 x3 ndash 1 lt16

3 3ή ndash 128 lt x lt 16

42 a y =ag

x +bg ndash ad

g2ln gx + d + c gne 0 gx + dne 0

b

x =ga y +

ad ndash bg

a2

ln ay + b + c a ne 0 ay + b ne 0

43 y = ndash x2y

x2+ 2y2 = krArr 44 y = rArr4yx y = k x4

45 x x ndashk 2 + y2 ndash k 2 + 1 = 0 k2 gt 1 46 y = rArrxy x2 ndash y2 = k

47 x ndash y = k x + y 3 48 x2 ndash 2ky=k2 kgt0

22

1 ex + (x 22) + (y 22) 2 όχι πλήρης

3 x3y ndash xy3 + (y 22) = c 4 t2 + z2 = c

5 ey ndash2 x2y ndash log x = c 6 όχι πλήρης

7 όχι πλήρης 8 x2 y ndash tanx + y2 = c

9 y = c ndash x 2 + xex ndash1 10 y = c ndash 3x x2 ndash 1ndash1

11 όχι πλήρης 12 sin x cosy + x2 ndash y2 = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 5

13 r = c ndash eθ sec θ 14 y = c + et t ndash 1 1 + et ndash1

15 x2 + x y2 ndash sin x + y ndash e y = c 16 όχι πλήρης

17 cos xyndash1 = c ή xyndash 1= c 18 όχι πλήρης

19 ln1 + xy

1 ndash xyndash 2x = c 20 x ey2

+ csc y cotx = c

21 x2 + y2 = c 22 x3

1 + ln y ndashy2

= c

23 3 x2y + y3 = 32 24 x2 + 2 xy ndash y2 = ndash16

25 x sin y + y2 = π2 26 x2 e2y + 2y = 4

27 y = x + 28 3x2 2 x lt 283 28 y = x ndash 24x3 + x2 ndash 8x ndash 16 1 21 2 4

29 xex y3 + ex ndash 6y3 = ndash5 30 2 cosx cosy = 1

31 exy2+ x4 ndash y3 = 2 32 y = e ndash x ex ndash1 ndash1 33 y = ndash2 tet + 2

ndash1

34 a Ν(x y) = xcos xy + g y b N x y = x exy ndash x 4 + g y όπου g αυθαίρετη

συνάρτηση του y

36 a k = 3 x2 y2 + 2x3y = c b k = 1 x2 + e2 xy = c

37 2x = y3 + cy 38 xy ndash x ndash 2y = c

39 x3 y + ey = c 40 x + arc tan y xndash1 = c

23

1 μ = yndash 4

x2y

ndash3ndashy

ndash1= c y equiv 0 2 μ = x

ndash1 xy ndash ln x ndash 2

ndash1y

2= c x equiv 0

3 μ =1

x3 y3 ndash

1

2x2 y2+

3

2y2 = c xequiv0 y equiv 0

4 μ t = et y t = ndash et plusmn e2t + 2 cendash t 1 21 2

5 μ t = endash at cosy a = y t = arcsin c ndasht et

6 μ = ty lny

t+ ty = c t equiv 0 y equiv 0

7 μ=t

ndash2y

ndash3

t2

y2

+ 1t = c t equiv 0 y equiv 0

6 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

8 μ=ndash tndash1

yndash2

2ln ty + ty = c t equiv 0 y equiv 0

9 μ = xy x3y + 3x2 + y3 = c

10 μ = x2

ndash y2 ndash1

x ndash y + ln x + y ndash ln x ndash y = c y = x y = ndashx

11 μ = endash y 2 xex ndash y + y2 = c 12 μ = x y ndash 4 34 3 y ndash x + 1 3 = c xy xequiv 0 y equiv 0

13 μ = xy x + y + 1 ndash 1 x + y + 1 3 = c xy xequiv 0 y equiv 0

14 μ = x 6x2 y2 + 8x3y + 3x4 = c x0

15 μ = x2

+ y2 1

y + arc tanyx = c y equiv 0

16 μ = xa yb a = b = x7 y2 ndash x3 y9 = c x 0 y 0

17 μ = eax

eby

a = b = x2

ndash 2xy endash x + 2y

= c xequiv0

18 μ = xa

ebx

a = b = x

2

2ye

6x+

e6x

96x ndash1 = c xequiv0

19 μ = xa

yb

a = b = x6

y4

+ x5

y7

= c xequiv0 yequiv0

20 μ = xa yb a = b = x3 y4 = 1296

21 μ = yaexb

a = b = y3ex2

= 125 e9

22 μ = xa

yb

ex

g

a = b = g = yx2

ex

2

= 12 e4

23 μ = x x2 y3 ndash 2 x2 = ndash 1143

24 μ = exbb = xy ex2

= ndash 2e 25 μ = σταθερά 4x2 + 3y2 = 247

26 μ = xa eb a = b = 3 x23 yex = ndash 21e

24

1 y = endashx

arc tanndash1

ex

+ cendashx

2 y = 1 + x2 ndash1

ln sin x + c 1 + x2 ndash1

3 y = x2 endashx + x2 ndash 2x + 2 + cendashx 4 y = x2 csc x + c csc x

5 xy sin x = sin x ndash xcosx + c 6 y = x3

+ c ln x3

7 y =cx +

3cos2x

4x+

3

2sin 2x 8 y = arc tan x + c 1 + x

2 ndash2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 7

9 y = 1 ndash

12

endash2x ndashinfin lt x lt infin

10 y =

qp + y0 ndash

qp endashp x ndash x

0 xisinΙR

11 y = x2

ndash 52x

0 lt x lt infin

12 y = x + x02 y0 ndash x0 x2 0 lt x lt infin αν x0 gt 0 και ndashinfin lt x lt 0 αν x0 lt 0

13 y = x + ln xx0 + x0 y0 ndash x0

2 x οltxltinfin αν x0gt0 και ndashltxlt0 αν x0lt0

14 y = a + ln 1 + x endash x

ndash1 lt x lt infin

15 y = 14

x2 ndash 13

x + 12

+ 112

xndash2 lt x lt infin 0

16 y = tanx + a sec x ndashπ22 lt x lt π22

17 y(x)0 όταν λlt0 y(x) y0 όταν λ=0 και y(x) +infinόταν λgt0

19 y(x) = x2 όταν x0 και y(x) = x2ndash2x3 όταν xge0

20 y =

x ndash 13

6+ c1 ln x ndash 1 + c2 21 y x =

ex x lt 02ex ndashx ndash1 x ge 0

22 y x =endash x x + 1 0 le x lt 2

2endash x + endash2 23 y x =

12

1 ndash endash 2x 0 le x le 1

12

e2 ndash1 endash2 x 1 lt x

24 y x =endash2x 0 le x le 1

endash x + 1 1 lt x 27 ναι αν y0 28 ναι 29 ναι

30 όχι 31 ναι 32 ναι 33 ναι 34 ναι 35 y = x2 ln x + cx2

36 y = sin x + c cosx 37 y =12

ndash1x +

c

x2

38 y =x2 x ndash 1

2+ c 1 ndash x 39 y = endashx +

cendashx

x

40 3y4 x8 ndash 2x6 = c 41 x2 yndash2 + 2 ex = c

42 ndash 3 x2 yndash 2 32 3 ndash 4x = c 43

u = xyndash 2 y = 23 + cx2

44 y =plusmn k l + cke ndash2kxndash 1 2

45 y = plusmn σ 12x 2 σ s ds + c

xndash12

όπου σ x = exp 2Γ sin x + 2Tx

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 5: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 5

13 r = c ndash eθ sec θ 14 y = c + et t ndash 1 1 + et ndash1

15 x2 + x y2 ndash sin x + y ndash e y = c 16 όχι πλήρης

17 cos xyndash1 = c ή xyndash 1= c 18 όχι πλήρης

19 ln1 + xy

1 ndash xyndash 2x = c 20 x ey2

+ csc y cotx = c

21 x2 + y2 = c 22 x3

1 + ln y ndashy2

= c

23 3 x2y + y3 = 32 24 x2 + 2 xy ndash y2 = ndash16

25 x sin y + y2 = π2 26 x2 e2y + 2y = 4

27 y = x + 28 3x2 2 x lt 283 28 y = x ndash 24x3 + x2 ndash 8x ndash 16 1 21 2 4

29 xex y3 + ex ndash 6y3 = ndash5 30 2 cosx cosy = 1

31 exy2+ x4 ndash y3 = 2 32 y = e ndash x ex ndash1 ndash1 33 y = ndash2 tet + 2

ndash1

34 a Ν(x y) = xcos xy + g y b N x y = x exy ndash x 4 + g y όπου g αυθαίρετη

συνάρτηση του y

36 a k = 3 x2 y2 + 2x3y = c b k = 1 x2 + e2 xy = c

37 2x = y3 + cy 38 xy ndash x ndash 2y = c

39 x3 y + ey = c 40 x + arc tan y xndash1 = c

23

1 μ = yndash 4

x2y

ndash3ndashy

ndash1= c y equiv 0 2 μ = x

ndash1 xy ndash ln x ndash 2

ndash1y

2= c x equiv 0

3 μ =1

x3 y3 ndash

1

2x2 y2+

3

2y2 = c xequiv0 y equiv 0

4 μ t = et y t = ndash et plusmn e2t + 2 cendash t 1 21 2

5 μ t = endash at cosy a = y t = arcsin c ndasht et

6 μ = ty lny

t+ ty = c t equiv 0 y equiv 0

7 μ=t

ndash2y

ndash3

t2

y2

+ 1t = c t equiv 0 y equiv 0

6 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

8 μ=ndash tndash1

yndash2

2ln ty + ty = c t equiv 0 y equiv 0

9 μ = xy x3y + 3x2 + y3 = c

10 μ = x2

ndash y2 ndash1

x ndash y + ln x + y ndash ln x ndash y = c y = x y = ndashx

11 μ = endash y 2 xex ndash y + y2 = c 12 μ = x y ndash 4 34 3 y ndash x + 1 3 = c xy xequiv 0 y equiv 0

13 μ = xy x + y + 1 ndash 1 x + y + 1 3 = c xy xequiv 0 y equiv 0

14 μ = x 6x2 y2 + 8x3y + 3x4 = c x0

15 μ = x2

+ y2 1

y + arc tanyx = c y equiv 0

16 μ = xa yb a = b = x7 y2 ndash x3 y9 = c x 0 y 0

17 μ = eax

eby

a = b = x2

ndash 2xy endash x + 2y

= c xequiv0

18 μ = xa

ebx

a = b = x

2

2ye

6x+

e6x

96x ndash1 = c xequiv0

19 μ = xa

yb

a = b = x6

y4

+ x5

y7

= c xequiv0 yequiv0

20 μ = xa yb a = b = x3 y4 = 1296

21 μ = yaexb

a = b = y3ex2

= 125 e9

22 μ = xa

yb

ex

g

a = b = g = yx2

ex

2

= 12 e4

23 μ = x x2 y3 ndash 2 x2 = ndash 1143

24 μ = exbb = xy ex2

= ndash 2e 25 μ = σταθερά 4x2 + 3y2 = 247

26 μ = xa eb a = b = 3 x23 yex = ndash 21e

24

1 y = endashx

arc tanndash1

ex

+ cendashx

2 y = 1 + x2 ndash1

ln sin x + c 1 + x2 ndash1

3 y = x2 endashx + x2 ndash 2x + 2 + cendashx 4 y = x2 csc x + c csc x

5 xy sin x = sin x ndash xcosx + c 6 y = x3

+ c ln x3

7 y =cx +

3cos2x

4x+

3

2sin 2x 8 y = arc tan x + c 1 + x

2 ndash2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 7

9 y = 1 ndash

12

endash2x ndashinfin lt x lt infin

10 y =

qp + y0 ndash

qp endashp x ndash x

0 xisinΙR

11 y = x2

ndash 52x

0 lt x lt infin

12 y = x + x02 y0 ndash x0 x2 0 lt x lt infin αν x0 gt 0 και ndashinfin lt x lt 0 αν x0 lt 0

13 y = x + ln xx0 + x0 y0 ndash x0

2 x οltxltinfin αν x0gt0 και ndashltxlt0 αν x0lt0

14 y = a + ln 1 + x endash x

ndash1 lt x lt infin

15 y = 14

x2 ndash 13

x + 12

+ 112

xndash2 lt x lt infin 0

16 y = tanx + a sec x ndashπ22 lt x lt π22

17 y(x)0 όταν λlt0 y(x) y0 όταν λ=0 και y(x) +infinόταν λgt0

19 y(x) = x2 όταν x0 και y(x) = x2ndash2x3 όταν xge0

20 y =

x ndash 13

6+ c1 ln x ndash 1 + c2 21 y x =

ex x lt 02ex ndashx ndash1 x ge 0

22 y x =endash x x + 1 0 le x lt 2

2endash x + endash2 23 y x =

12

1 ndash endash 2x 0 le x le 1

12

e2 ndash1 endash2 x 1 lt x

24 y x =endash2x 0 le x le 1

endash x + 1 1 lt x 27 ναι αν y0 28 ναι 29 ναι

30 όχι 31 ναι 32 ναι 33 ναι 34 ναι 35 y = x2 ln x + cx2

36 y = sin x + c cosx 37 y =12

ndash1x +

c

x2

38 y =x2 x ndash 1

2+ c 1 ndash x 39 y = endashx +

cendashx

x

40 3y4 x8 ndash 2x6 = c 41 x2 yndash2 + 2 ex = c

42 ndash 3 x2 yndash 2 32 3 ndash 4x = c 43

u = xyndash 2 y = 23 + cx2

44 y =plusmn k l + cke ndash2kxndash 1 2

45 y = plusmn σ 12x 2 σ s ds + c

xndash12

όπου σ x = exp 2Γ sin x + 2Tx

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 6: 978-960-7182-07-4

6 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

8 μ=ndash tndash1

yndash2

2ln ty + ty = c t equiv 0 y equiv 0

9 μ = xy x3y + 3x2 + y3 = c

10 μ = x2

ndash y2 ndash1

x ndash y + ln x + y ndash ln x ndash y = c y = x y = ndashx

11 μ = endash y 2 xex ndash y + y2 = c 12 μ = x y ndash 4 34 3 y ndash x + 1 3 = c xy xequiv 0 y equiv 0

13 μ = xy x + y + 1 ndash 1 x + y + 1 3 = c xy xequiv 0 y equiv 0

14 μ = x 6x2 y2 + 8x3y + 3x4 = c x0

15 μ = x2

+ y2 1

y + arc tanyx = c y equiv 0

16 μ = xa yb a = b = x7 y2 ndash x3 y9 = c x 0 y 0

17 μ = eax

eby

a = b = x2

ndash 2xy endash x + 2y

= c xequiv0

18 μ = xa

ebx

a = b = x

2

2ye

6x+

e6x

96x ndash1 = c xequiv0

19 μ = xa

yb

a = b = x6

y4

+ x5

y7

= c xequiv0 yequiv0

20 μ = xa yb a = b = x3 y4 = 1296

21 μ = yaexb

a = b = y3ex2

= 125 e9

22 μ = xa

yb

ex

g

a = b = g = yx2

ex

2

= 12 e4

23 μ = x x2 y3 ndash 2 x2 = ndash 1143

24 μ = exbb = xy ex2

= ndash 2e 25 μ = σταθερά 4x2 + 3y2 = 247

26 μ = xa eb a = b = 3 x23 yex = ndash 21e

24

1 y = endashx

arc tanndash1

ex

+ cendashx

2 y = 1 + x2 ndash1

ln sin x + c 1 + x2 ndash1

3 y = x2 endashx + x2 ndash 2x + 2 + cendashx 4 y = x2 csc x + c csc x

5 xy sin x = sin x ndash xcosx + c 6 y = x3

+ c ln x3

7 y =cx +

3cos2x

4x+

3

2sin 2x 8 y = arc tan x + c 1 + x

2 ndash2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 7

9 y = 1 ndash

12

endash2x ndashinfin lt x lt infin

10 y =

qp + y0 ndash

qp endashp x ndash x

0 xisinΙR

11 y = x2

ndash 52x

0 lt x lt infin

12 y = x + x02 y0 ndash x0 x2 0 lt x lt infin αν x0 gt 0 και ndashinfin lt x lt 0 αν x0 lt 0

13 y = x + ln xx0 + x0 y0 ndash x0

2 x οltxltinfin αν x0gt0 και ndashltxlt0 αν x0lt0

14 y = a + ln 1 + x endash x

ndash1 lt x lt infin

15 y = 14

x2 ndash 13

x + 12

+ 112

xndash2 lt x lt infin 0

16 y = tanx + a sec x ndashπ22 lt x lt π22

17 y(x)0 όταν λlt0 y(x) y0 όταν λ=0 και y(x) +infinόταν λgt0

19 y(x) = x2 όταν x0 και y(x) = x2ndash2x3 όταν xge0

20 y =

x ndash 13

6+ c1 ln x ndash 1 + c2 21 y x =

ex x lt 02ex ndashx ndash1 x ge 0

22 y x =endash x x + 1 0 le x lt 2

2endash x + endash2 23 y x =

12

1 ndash endash 2x 0 le x le 1

12

e2 ndash1 endash2 x 1 lt x

24 y x =endash2x 0 le x le 1

endash x + 1 1 lt x 27 ναι αν y0 28 ναι 29 ναι

30 όχι 31 ναι 32 ναι 33 ναι 34 ναι 35 y = x2 ln x + cx2

36 y = sin x + c cosx 37 y =12

ndash1x +

c

x2

38 y =x2 x ndash 1

2+ c 1 ndash x 39 y = endashx +

cendashx

x

40 3y4 x8 ndash 2x6 = c 41 x2 yndash2 + 2 ex = c

42 ndash 3 x2 yndash 2 32 3 ndash 4x = c 43

u = xyndash 2 y = 23 + cx2

44 y =plusmn k l + cke ndash2kxndash 1 2

45 y = plusmn σ 12x 2 σ s ds + c

xndash12

όπου σ x = exp 2Γ sin x + 2Tx

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 7: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 7

9 y = 1 ndash

12

endash2x ndashinfin lt x lt infin

10 y =

qp + y0 ndash

qp endashp x ndash x

0 xisinΙR

11 y = x2

ndash 52x

0 lt x lt infin

12 y = x + x02 y0 ndash x0 x2 0 lt x lt infin αν x0 gt 0 και ndashinfin lt x lt 0 αν x0 lt 0

13 y = x + ln xx0 + x0 y0 ndash x0

2 x οltxltinfin αν x0gt0 και ndashltxlt0 αν x0lt0

14 y = a + ln 1 + x endash x

ndash1 lt x lt infin

15 y = 14

x2 ndash 13

x + 12

+ 112

xndash2 lt x lt infin 0

16 y = tanx + a sec x ndashπ22 lt x lt π22

17 y(x)0 όταν λlt0 y(x) y0 όταν λ=0 και y(x) +infinόταν λgt0

19 y(x) = x2 όταν x0 και y(x) = x2ndash2x3 όταν xge0

20 y =

x ndash 13

6+ c1 ln x ndash 1 + c2 21 y x =

ex x lt 02ex ndashx ndash1 x ge 0

22 y x =endash x x + 1 0 le x lt 2

2endash x + endash2 23 y x =

12

1 ndash endash 2x 0 le x le 1

12

e2 ndash1 endash2 x 1 lt x

24 y x =endash2x 0 le x le 1

endash x + 1 1 lt x 27 ναι αν y0 28 ναι 29 ναι

30 όχι 31 ναι 32 ναι 33 ναι 34 ναι 35 y = x2 ln x + cx2

36 y = sin x + c cosx 37 y =12

ndash1x +

c

x2

38 y =x2 x ndash 1

2+ c 1 ndash x 39 y = endashx +

cendashx

x

40 3y4 x8 ndash 2x6 = c 41 x2 yndash2 + 2 ex = c

42 ndash 3 x2 yndash 2 32 3 ndash 4x = c 43

u = xyndash 2 y = 23 + cx2

44 y =plusmn k l + cke ndash2kxndash 1 2

45 y = plusmn σ 12x 2 σ s ds + c

xndash12

όπου σ x = exp 2Γ sin x + 2Tx

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 8: 978-960-7182-07-4

8 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

25

1 y = x + c ndash x ndash1 2 y = xndash1 + 2x c ndash x2 ndash1

3 y = sin x + c cos x ndash 12 sin x 4 y = x ndash 2 + cendashx ndash 1 + 1

5 y = 2 + cendash 2x2 ndash1+ x 6 y = 1 + 1 ndash t + cendasht ndash 1

7 y = et + cendash 3t ndash e ndasht 2ndash 1

8 y= 1 ndash tcendasht 1 ndash cendasht ndash1

9 y = t + t c ndasht5

5

ndash1

10 y =2x +

1

cxndash3 ndash x 4x 4

11 y = ndashex + cendashx ndash 1 ndash1 12 y = ndash x2 + 2x2 ex2

ex2+ c

ndash1

13 y =1

cosx +3 cos2x

c ndash cos3x 14 y =

1x + 2 cx3 ndash x ndash1

15 y = x + 2x cx2

+ 1ndash1

16 y = x + cendashx ndash 1 ndash1

17 y = 1 + x 1 ndash x + cendashx

23 pndash 1 my2 + ny + pndash1

dx = C ndash f 0 x dx

25 Συνθήκη b ndash2 f1b f2ndash1 bndash2 = ndash k + 1

k

26

1 βαθμού 3 2 βαθμού 2

3 όχι 4 βαθμού 0

5 βαθμού ndash2 6 βαθμού ndash

13

7 όχι 8 βαθμού 3

11 x + yln x = cy 12 ln x2 + y2 + 2tanndash1 y

x = c

13 4x = y ln y ndash c2

14 y9 = c x3 + y3 2

15 yx

2= 2ln x + c 16 xcos

yx = c

17 y + x = cx2 ey xy x 18 y2 ndash cx = y y2 ndash x2

19 ysinyx = c 20 y = c 1 + ln

yx

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
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  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 9: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 9

21 2 ex yx y + ln y = c 22 ln x

2ndashe

y xy x sinyx + cos

yx = c

ndash

23 y = 2x tan 2ln x + tanndash1

ndash05 24 y3 + 3x3 ln x = 8x3

25 ln x = ey xy x ndash 1 26 4x lnyx + x ln x + y ndash x = 0

27 3x3 23 2 ln x + 3x1 21 2y + 2y3 23 2 = 5x3 23 2 28 x + y ln y + x = 0

29 ln y = ndash2 1 ndash x y1 21 2 + 2 30

y2 ln x = 2y2 + xy ndash x2

32 x ndash 4 = c 1 ndash4 y + 1

x ndash 4

ndash1 41 4 33 3 x ndash 2 2 ndash 2 x ndash 2 y + 3 ndash y + 3 2 = c

34 ndash8ln x ndash 2y + 4 + 2x ndash 6y = c 35 ndash2

15tanndash1

2 y ndash 1 ndash 3 5

5 x + 1= ln x + c

36 516

ndash2u + z + 13128

ln 16u ndash 8z = ndash 12

u+c

37 5x ndash 3

t ndash 4

2ndash 2

x ndash 3

t ndash 4+ 2

ndash1 21 2= c t ndash 4

38 (i) x + y ndash 3 3 = c 2x + y ndash 4 2 (ii) x + 2y + c = 3ln x + y + 2

(iii) x ndash y ndash 4 3 = c x + y ndash 2 (iv) 2y ndash x + 3 2 = c y ndash x + 2

27

1 c2 + cx2 = 2y 8y = ndashx4 2 12y = c c + 4x3 3y = ndashx6

3 2c3 x3 = 1 ndash 6c2y 2y = x2 4 x3 y + c2 + c = 0 4x6y = 1

5 x2 = c y ndash c y = plusmn2x 6 xy = c 3cx ndash 1 12x2y = ndash1

7 xc2 + x ndash y c + 1 ndash y = 0 x + y 2 = 4x 8 3xy = c xc2 ndash 3 9x3 y2 = 4

9 2c3y = c4 x2 + 12 3y2 = plusmn8x3 10 p3 x + 2p 2 = c y = 3xp + 5p2

11 x2 = cpndash4 34 3 ndash 2pndash1 y = xp + x3 p2 12 y = cx + 1 ndash ln c y = 2 + tanx

13 y = cx ndash c3 27y2 = 4x3 14 y = cx ndash ec y = x ln x ndash x

15 y = cx + c2 + 1 y = 1 ndashx2 4 16 y = cx ndash c2 32 3 27x2y + 4 = 0

17 y = cx + 1 + c2 y = 1 ndash x2 y gt 0 18 y3 = cx + 2c23

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 10: 978-960-7182-07-4

10 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

19 27 ay2 ndash 16x3 y2 + 16a2 x 9ay2 ndash 4x3 c ndash 128a3 x2 c2 ndash 64a4 c3 = 0

20 y2 4y ndash 3x2 + 6x 2x2 ndash 3y c + 9c2 = 0

28

1 siny = x2 ndash 2x + 2 + cendashx 2 y + 1 = x + 1 + c x + 1

3 cosy = 1 ndash cendashcosx 4 y = c + x2 + y2

5 endash 2x

= 2y2

ln y ndashc y ne 0 y = y x equiv 0 6 exy 2+ x4 ndash y3 = c

7 3 x2 + 2y ndash 11 31 3 = 2x + c 8 x

2e2y = 2x lnx ndash 2x + c

9 ndash e yx4

= x2 + c 10 ln tany = x + cxndash1

11 x3 y3 = 2x3 ndash 9ln x + c 12 ey = ndashendashx cosx + cendashx

29

1 y2 + yx = cx3 2 x3

3+ xy + ey = c

3

1y = ndashx

e2t

t2

x

dt + cx y = 0 4 2 arcsinyx ndash ln x = c

5 x + 2ln y ndash 2ln ln x = c 6 u = x + y x + y = tan x + c

7 yndash2 = ce2x2ndash 4x2 ndash 2 8 y = ux y = x sin c plusmn x2 + y2 2 x 0ltgt

9 sinx = ceay ndashy2

a +2y

a2+ 2

a3

10 Oμογενής 1 ndash cx ey xy x = cx

11 u = xy χωρ μετ xy = ecx

12 Bernoulli ολ παρ xndash2 yndash2 x2 ndash y = c xy

13 πλήρης 3xy2 + x2y + 3y + x2 = c 14 πλήρης x2 y2 ndash 2 x + y = c

15 πλήρης x3 + y3 + 3 x2y + y + 3x = c 16 Bernoulli 2x2 = 2y ndash 1 + cendash2y

17 y = ux2 x2 + y2

= cy 18 Bernoulli y2 = cx ndash 3x2

19 Oλ παρ xndash2 yndash2 x3 + y3 = c xy 20 u = xy y2 = cexp xy ndash xndash1yndash1

21 Oμογενής πλήρης y3 + 4x3 + 3x2y = c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 11: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 11

22 Bernoulli xy3 = x2 + c 23 Oλοκλ παραγ x2 + y2 x2 + y2 2 2y2 ndash x2 = c

24 Oλοκλ παραγ xndash 1

yndash 2

y3

+ yln xy ndash x = cy

25 χωρ μετ πλήρης y4 + y2 = x4 + x2 + c 26 u = ey ey = 1 + cendashex

27 z = x + y ndash 2 y = 2 ndash x + tan x ndash 1 + π44

28 Πλήρης xy ndash 3x22 ndash 8x ndash 2y = c 29 χωρ μετ y = sin x +π

4

30 γραμμική ως προς y y = x ndash 1 3 6 + cln x ndash 1 + k

31 y1 = ndash2 y = ndash2 + (cendashxndash1)ndash1 32 y1 =ex

y y = ex + 2 cendash3x ndash ex ndash1

39 M =x2y + c 40 M =2xndash1 yndash3 ndash3

2x2 yndash4 + c

41 A =3

2 2x3 + 9x2y + 12y2 = c 42 A =ndash2 2x2 ndash 2y2 ndash x = c xy2

45 x y2 + 1 = cy 46 x3y 2x ndash 3y = c

47 2x3 ndash y 2 = cyx6 48 y2 c ndash x = x3

49 x + 2y ndash 1 2 = 2y + c 50 5x2 y3 = x5 + 4

51 2y2 = x2 3x ndash 1 52 x ndash 3 2 x ndash 3y = c

53 x2 + x ndash 3xy ndash y2 + 4y = c 54 b2y = c1 ebx ndash a bx ndash a ndash cb

55 ln x ndash y + 1 = c ndash x

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 12: 978-960-7182-07-4

2 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17 r = c eϕ 18 ϕ2 = ndash 4 ln r + c

19 r = c sin ϕ 20 r2 = c cos 2ϕ

21 r = c cscϕ 22 c r = eϕ2 2

23 e1r = c sec ϕ + tan ϕ 24 r = 2 sin ϕ + c cos ϕ

25 r2 = c sin 2ϕ 26 rn cos nϕ = c

27 r= eplusmn c ndash ϕ2

29

23

tanndash 1 y x + ln c x2 + y2 = 0

30 ln x + c = tanndash 1 y

x ndash 12

ln 1 +y2

x2

31

3 + tanndash1 y

1 + xndash 1

2ln x + 1 2 + y2 = c

32 ndash y + 2 ln 1 + y = x + c 33 y = 43

ln 3 x ndash 1 + 13

x + c

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError 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Page 13: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 3

32 Προβλήματα

2 y t = y0 + t z0_ t _ s

0

t

g s y s ds

3 y t = y0 cos microt+ micro

_1 z0 sin microt + micro_1 sin micro t _ s

0

tg s y s ds

22 ναι 23 ναι

24 ναι 25 όχι

26 όχι 27 ναι

28 ναι 29ναι

37 y x = _ x22

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 14: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 4

41 Προβλήματα

1 y = c2 ec1x 2 A = c1e

ndash1 u1 u + c2

3 y = tan x 2 4 y = e2 x ndash 1

5 y =x

2

2+

3

2 6 y = 2 + ln

x2

ndash 3

2με τασχηματισμός z = y prime

ndash 1

7 x5

5+ 2c1

x3

3+ c1

2x + c2 y equiv c 8 y = plusmn 13

2x + c13 23 2

+ c2

9 x = c1 + c2y + y3 6 y equiv c

10 y = c1

c2 ec1x ndash 1

c2 ec1x + 1

ή y = 2 x + cndash 1y = c1 tan c2 ndash c1 ή x

2

11 y = c1 + c2x

3+

x4

4ndash

3x5

10+

x6

18

12 y ndash ln ey

+ c1 = c1x + c2 για c1 ne 0 ή y = ndash ln c ndash x για x lt c ή y equiv c

13 x = c2 ndash c1y + 1 + c1

2 ln y + c1 ή y equiv c

14 y = plusmn 23 x ndash 2c1 x + c1 + c2 ή y equiv c 15 x + c2 = plusmn 23 y ndash 2c1 y + c11 21 2

16 y ln y ndash y + c1y + x = c2 ή y = c 17 ey

= x + c22

+ c1

23

a2

ex ax a + endash x ax a = a cos h xa αλυσοειδής καμπύλη

42 Προβλήματα

10 3x2

ndash 8x + 2 11 0 12 r2 ndash 4r + 3 erx

16 όχι 17 όχι 18 ναι (ndash 1)

19 ναι IR 20 ναι (0 ) 21 ναι (0 )

22 ναι (0 ) 23 ναι (ndash +) 24 a 2 b x sin x + cos x

25 α 3x2 ndash x 3 b 6x + 3x2 ndash 2x3 c 3x2

+ 2x3

d 6x + 3x2

ndash 2x3

e D2

+ D ndash 2 f 6x + 3x2

ndash 2x3

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
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  • Κεφάλαιο 10
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StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true 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Page 15: 978-960-7182-07-4

16 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

43 Προβλήματα

1 ndash b sin at sin bt + a cos at cos bt 2 0

3 b ndash a e a + b t 4 ndash b e2at 5 t

6 W 1 = 5e3

W ndash 1 = 5 endash 3

W 3 = 5 endash 9

W x = 5 e3x

8 y Prime + y primendash 6y = 0 9 x2

y Primendash 3 xy prime+ 4y = 0

10 xyPrime ndash y prime+ 4 x3y = 0 11 yPrime + y prime

2+ 1 = 0

12 yyPrime + y prime2

= 2 13 yyPrime + yprime3= 0

14 ναι 15 όχι 17 ναι 18 ναι 19 ναι 20 ναι

22 J = (ndash2 2) 23 J = (0 +)

31 y = c1 e2x

+ c2 xe2x

32 y = c1 x + c2 x ln x

33 y = c1 e2x

+ c2 2x2

+ 2x + 1 34 y = c1 x cosx + c2 x sinx

35 y = c1 x2

+ c2 2x + 1 36 y = c1x3

+ c2xndash 3

37 y = c1 x2

sinx + c2 x2

cosx 38 y = c1 x3 23 2 + c2 x

3 23 2 ln x

39 y = c1 xndash 2

+ c2 xndash 3

41 y x = x2 + 1 c1 + c2 x x2 + 1ndash1

+ tanndash 1 x

42 y x = c1x

x + 12

+ c2x

2ndash 2x lnx ndash 1

x + 12

44 Προβλήματα

1 y = c1 ex 2x 2 + c2 e

4x 34x 3 2 y = c1 endash 1 + 2 x

+ c2 endash 1 ndash 2 x

3 y = c1 e3x

+ c2 endash 2 + 2 x

+ c3 endash 2 ndash 2 x

4 y = c1 + c2 ex

+ c3 e2x

+ c4 endash 2x

5 y = c1 ea x

+ c2 endash a x

+ c3 cos a x + c4 sin a x 6 y = c1 endash 6kx

+ c2 e2kx

7 y = c1 + c2x + c3 x2

+ c4 x3

8 y = c1 + c2x + c3x2

e2x

9 y = c1 + c2x + c3x2+ c4e

ndash 3x 10 y = c1 + c2x e2x

+ c3 + c4x endash 3x

11 y = c1 + c2x + c3 cos 2x + c4 sin 2x 12 y = 2e4x

+ endash 3x

13 y = 3x + 2 endash 3x 14 y = 13x + 3 endash 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 16: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 17

15 y = e2x

sin 5x 16 y = endash 3x

4 sin 2x + 3 cos 2x

17 y = 3endash x 33 sin

2x

3+ 2 cos

2x

3 18 y = e

xndash 2e

2x+ e

3x

19 y = 32 endash x ndash 23 e2x + 6 xe2x 9 20 y = 9 ex 3x 3 + 4x ndash 9 endash x 16

21 y =14

e2x ndash π

sin 4x 22 y = sin 2 x + π

23 B = es

+ endash s

+ cos s ndash sin s 24 y = cos 3 x ndash 3 ndash1

3sin 3 x ndash 3

45 Προβλήματα

2 y = c1e3x+c2e

ndashxndash4xendashx(2x+1) 3 y = c1 + c2x +12

x2

ln x ndash3

4x

2e

ndash x

4 y = c1 cos 2x + c2 sin 2x + 12

x sin 2x+ 14

cos 2x ln cos 2x endash x

5 y = c1 + ln 1 + endash x

ex

+ c2 + ln 1 + endash x

e2x

6 u t =c1et+c2endash t +c3endash 2 t + 16 t

0

t2endash 2 t ndash s ndash 3endash t ndash s +et ndash s f s ds

7 u t = 2 ndash3 c endash t

+ cet+ 2c ndash 1 e

ndash 2t+

+

1

6 0

t

2endash 2 t ndash s

ndash 3endash t ndash s

+ et ndash s

f s ds c σταθερά

8 u t = c1 tndash 1

ndash 1 + c2 t ndash1 + 12 1

t 12

tndash 1 + 12

tsndash 2 ndash 2sndash 1 f s ds

9 R(x)=(c1+c2xndashlnx)endash3x x 0

10 y x = c1 cos x + c2 sin x + tan x sin x ndash12

tan2x cos x

11 y x = c1x + c2x

2ndash x ln x ndash

12

x ln x2

12 N(x) = c1+c2cosx+c3sinx+lnsecx+tanxndashxcosx+sinxlncosx

13 y = ex

c1 cos 3 x + c2 sin 3 x +sin x

2

14 y(x) = c1e2x+c2e3x+ex[(Ax+B)cos2x+(Cx+D)sin2x]

15 y(x) = c1cosx+c2sinx+x[(A0+A1x+A2x2)cosx+(B0+B1x+B2x2)sinx]

16 y = 2 cos 2t ndash23

sin 2t +13

sin t

17 y =46

50ndash

135x

50+

x2

2e

3x+

4

50cos x ndash

3

50sin x

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 17: 978-960-7182-07-4

18 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

18 y s = ndash 2endash s

+ 56

endash 2 s

ndash 13

es+ 3

2 19 y(t) = 2etndashendashtndash2endashtsint

20 y(t) = 8e2tndash24et+4t2+12t+14+2endasht

21 y = c1 cos 2t + c2 sin 2t + 14

t2 ndash 18

+ 364

8t2 sin 2t ndash 3 sin 2t + 4t cos 2t

22 y t = endash t + 7 et cos 2t + et sin 2t 8

23 y t = ndash73

300e

2tndash

51

50te

2t+

13

endash t

+ t +12

t2

+3

25sin t +

4

25cos t +

3

4

24 y t = 31128

e2t + 97128

endash 2 t + 132

te2t ndash 116

t2 e2t + 112

t3 e2t

27 y(t) = c1cost+c2sint+z(t) όπου z(t) = t για t[0 π] z t = ndashπ2

t sin t για

t(π 2π) και z(t) για t 2π

29 φ(t) = (c1 - t) cos(kt) + (c2 - t) sin(kt)

30 y x = c1 cos λx + c2 sin λx +kn

λ2ndash n

2p

n = 1

Nsin n πx

33 3x2 34 (i) y x =1

10cos x 3 sin x (ii) y x =

1

10sin x + 3 cos x

46 Προβλήματα

1 y = 12

x5 endash2x 2 y =12

e5x

3 y = 2x3+9x2+40x+73 4 y = ndash1

60x

5ndash

13

x3

ndash 2x

5 y = ndashx10ndash1512x4 6 y = x4+4x3+24x2+69x+117

7 y = ndash2x3ndash5x2ndash10xndash10 8 y = ndashe2x(x3+6x)

9 y =1

8e

2x4x

3ndash 2x

2ndash 18x ndash 25 10 y = 2e

ndash 2xx

2+ 4x + 6 +

13

e2x

11 y = x3ndash1 12 y =12

x ln x ndash 1

13 y =1

484x

3ndash 6x

2+ 6x ndash 3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
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  • Κεφάλαιο 10
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StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError 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Page 18: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 19

47 Προβλήματα

4 yp =0

tK t s f s ds = Σ

j = 0

n ndash 1

2jπ

2j + 1 πK t s ds +

2nπ

t

K t s ds

όπου K t s = et ndash s 10s 10 sin t ndash s και 2nπ le t le 2n + 1 π

9 2π 199505 322

le T le 2π 201495 322

10 q t = ndash9

26e

ndash tcos 3t ndash

7

26e

ndash t sin 3t +

1

269 cos 2t + 6 sin 2t

12 ip =4

51cos 20t ndash

1

51sin 20t συχνότητα συντονισμού 5π

13 q = ndash

5

12sin 3t +

5

4sin t i = ndash

5

4cos 3t +

5

4cos t i0= 0

48 ANAKEΦAΛAIΩΣH ndash ΠPOBΛHMATA EΠIΣKOΠHΣHΣ

3 ndash m2le x le 0 ndash m

3tle xprime le m

3t

7 y t = 1 + ln t 41 + ln t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 19: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 5

52 Προβλήματα

1 r = 1 2 r=1

3 r = 23

4 r = 1

5 r =1 6 r = 1

7 r = + 8 r = 0

9 r = 1 10 r = 0

11 r = 12 r = 14

13 αποκλίνει στα plusmn1 14 συγκλίνει στα plusmn1

15 ndash 1

kndash1

2k + 1 Σ

k = 1

infinx2k+1= xndashsin x 16 1 + 2

2k + 1 Σ

k = 0

infinx2k+1= 1 + 2sin hx

17 x2k

kΣk = 1

infin 18 3ne3

nΣn = 0

infinx ndash 1

n

19 ndash 1

nndash1

nΣn = 1

infinx ndash 1

n 20 5+(ndash5) (x+1) + (x+1)2

21 ndash 1

n

3n+1Σn = 0

infinx ndash 2

n 22 ndash 1

nΣn = 0

infinx ndash 1

n

23 ndashx2n

2n+1Σn = 0

infin 24 ln 2 + 1

2x ndash 1

8x2 + 1

24x3 ndash 1

64x4 + 1

160x5

25 1 + π2

+ x ndash π2

ndash 2 x ndash π2

2+ 2

3x ndash π

24

26 2x2 ndash 13

x4

27 1 ndash x2 + 2x4 28 94

+ 3 x ndash 12

+ x ndash 12

2

29 sin 1 + cos 1 x ndash 12

sin 1 x2 ndash 16

cos 1 x3+ 124

sin 1 x4 + 1120

cos 1 x5

30 1 ndash 3x + 6x2 ndash 10x3 +15x4 ndash 21x5

31 ln 4

3+ 3

4x ndash 1

3ndash 9

32x ndash 1

3

2+ 9

64x ndash 1

3

3ndash 81

1024x ndash 1

3

4+ 243

5720x ndash 1

3

5

32 ndash 1nxn

2n ndash 2Σn = 4

infin 33

2n + 5 xn

n + 1Σn = 1

infin

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 20: 978-960-7182-07-4

22 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

34 2nndash1xn

n ndash 12

ndash 3Σ

n = 2

infin 35 2 n ndash 3

n ndash 1Σn = 3

infinxn

36 n ndash 2Σ

n = ndash 1

infinnxn

37 F x = 2a2 ndash a 1 + λa0 + 6a3 ndash 2a2 + λ ndash 2 a1 x

+ Σ

n = 2

infinn + 2 n + 1 an+2 + n + 1 an+1 + λ ndash n ndash n 2 an xn

38 F x = 2a2 + a1 + 6a3 x + Σn = 2

infinn + 2 n + 1 an+2 + n2an xn

40 1 + 2x + Σn = 2

infin2nndash1 + 1 + n xn 41 1 + 1

2x + Σ

n = 2

infin n + 1

n + 12

+ 2n xn

42 Σn = 1

infin1

2nndash n + 2

n+2xn 43 x + x2+ x3 + x4+ Σ

n = 5

infin1 + n ndash 2 xn

44 12

+ 13

x + Σn = 2

infin1

n + 2+ ndash 2

nndash1xn

53 Προβλήματα

1 όλα τα xIR ομαλά 2 όλα τα xIR ομαλά

3 ndash1 1 ΚΑ 4 01 ΚΑ

5 01 ΜΚΑ 6 0 ΚΑ

7 0 ΚΑ 1 ΜΚΑ 8 ndash1 ΜΚΑ

9 03 ΚΑ 10 0 ndash1 ΚΑ

11 2 ndash3 ΚΑ 0 ΜΚΑ 12 ndash1 0 ΟΜ 1 ΜΚΑ

13 ndash1 0 1 ΚΑ 14 ndash2 ndash1 0 1 ΚΑ

15 3i ndash3i ndash7 ΚΑ 16 2 plusmnnπ n = 1 2 hellip ΚΑ

17 ndash1 plusmnnπ n = 1 2 hellip ΚΑ 0 ΜΚΑ 18 ndash4 ndash2 4 ΚΑ 0 ΜΚΑ

19 ndash3 ndash1 1 2 ndashi i ΚΑ

20 y x = a0 Σk = 0

infink + 1 x2k + a1 Σ

k = 0

infin2k + 3

3x2k+1 ndash 1 lt x lt 1

21 y x = a0 Σk = 1

infin 4 ndash 1k

x ndash 13k

3k ndash 1 3k ndash4 k+ a1 x ndash 1 + 1

4x ndash 1

4 xisinIR

22 y x = a0 1 +

x2

2ndash Σ

k = 2

infinndash 1

k 1 3 5 2 k ndash 3

2kkx2k + a1x xisinIR

23 y x = a0 Σk = 0

infin1k

x2k xisinIR

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 21: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 23

24 y x = a0 1 ndash 2x2 + x43 + a1 x + Σk = 1

infin ndash 3 ndash 1 Ι 2k ndash 5

2k + 1 x2k+1 xisinIR

25 y x = a0 1 + Σk = 1

infin 1 middot 4 3 k ndash 5 3k ndash 2

3k x3k

+ a1 x + Σk = 1

infin 2 middot 5 3 k ndash 4 3k ndash1

3k + 1 x3k+1 xisinIR a3k+2= 0 k=0 1

26 y x = a0 Σk = 0

infinx2k + a1 Σ

k = 0

infinx2k+1 =

a0 + a1x

1 ndash x2 ndash 1 lt x lt 1

27 y x = a0 Σk = 0

infinndash 1

k x2k

k2k+ a1 Σ

k = 0

infin ndash 1k

x2k+1

2k + 1 xisinIR

28 y = a0 + a1 Σk = 0

infin x2k+1

2k + 1 3k xisin ndash 3 3

29 y x = a0 1 ndash 83

x2 + 827

x4 + a1 x ndash 12

x3 + 1120

x5 +

+ 9 Σ

k = 3

infin ndash 1 k 2k ndash 5 2

2k + 1 3kx2k+1 xisin ndash 3 3

30 y x = a0 1 + 23

x2 + 127

x4 + a1 x + 16

x3 + 1360

x5

+ 3 Σk = 3

infin ndash 1 k 2k ndash 5

2k + 1 3kx2k+1 xisinIR

31 y x = a0 Σk = 0

infin ndash 1k

x3k

k3k+ a1 Σ

k = 0

infin ndash 1k

x3k+1

1 4 Ι 3k + 1 xisinIR

32 y x = 1 + x +x2

2+

x4

24+ ndash 1 lt x lt 1

33 y x = 1 +x ndash 1 2

2+

x ndash 1 3

6ndash

x ndash 1 4

8+

x ndash 1 5

15ndash 0 lt x lt 2

34 y x = a0 1 ndashx2

2ndash

x3

6+

x4

12+

x5

20+ + a1 x ndash

x3

3ndash

x4

12+

x5

20+ xisinIR

35 y = 12

x ndash 1 + 12

x ndash 12

+ 16

x ndash 13

ndashx ndash 1

4

24+

x ndash 15

48+ 0 lt x lt 2

36 y x = 1 ndash x + x2 ndashx3

2ndash

x4

12ndash

x5

60ndash

x6

48+ ndash 1 lt x lt 1

37 y = a0 1 ndashx2

16ndash

x3

96+

5x4

1536+ + a1 x ndash

x3

24ndash

x4

192+

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 22: 978-960-7182-07-4

24 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

38 y x = 2 x ndash 1 + x ndash1 2 + 2 x ndash 1 3 +x ndash 1 4

6+ 0 lt x lt 2

39 y x = 1 ndashx ndash π 3

6+

x ndash π 5

120+

x ndash π 6

180+ xisinIR

40 y(0) = 0 y(0) = ndash2 y(4)(0) = 0

41 y(0) = 0 y(0) = ndasha0 y(4)(0) = ndash4a1

42 y(1) = 0 y(1) = ndash6 y(4)(1) = 42

43 ανώμαλο σημείο (Υπόδειξη να γίνει χρήση του Θεωρήματος Picard)

54 Προβλήματα

1 y x = a0 1 ndash1 + 1

2x2 +

1 + 1 1 ndash 2 1 + 34

x4 + + a1x

2 y x = a0 1 ndash 3x2 + a1 x ndash2 ndash 1 2 + 2

3x3 +

2 ndash 1 2 + 2 2 ndash 3 2 + 45

x5 + Ι

3 y x = x ndash 143

x3 + 215

x5

6 (i) y1 x b = 1 + Σ

k = 1

infin ndash b 4 ndash b 16 ndash b 2k ndash 22

ndashb

2k x2k

y2 x b = x + Σk = 1

infin 1 ndash b 9 ndash b 2k ndash 12

ndashb

2k + 1 x2k+1

(iv) T0 = 1 T1 = x T2 = 2x2ndash1 T3 = 4x3ndash3x T4 = 8x4ndash8x2+1

7 b = n (n+2) U0 = 1 U1 = 2x U2 = 4x2ndash1 U3 = 8x3ndash4x

55A Προβλήματα

1 y = c1+c2lnx 2 y = c1x3+c2x

-4

3 y = c1x12+c2x

-32 4 y = x[c1cos(lnx)+c2sin(lnx)]

5 y = c1cos 32

lnx + c2sin 32

lnx 6 y = c1+c2x2+c3x

2lnx

7 y = c1+c2lnx+c3(lnx)2 8 y = c1 xndash 1+ x1 21 2 c2cos 3

2lnx + c3sin 3

2lnx

9 y = c1 x3 + c2 xndash 2 ndash 16

10 y = c1 + c2lnx xndash 1 + 49

x

11 y = c1x + c2xndash 5 + x2 7 12 y = c1x-52+c2x

-3

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 23: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 25

13 y = c1x3cos(2lnx)+c2x

3sin(2lnx) 14 y = c1x+c2x-1cos(3lnx)+c3x

-1sin(3lnx)

15 y = c1x-2+c2x

-2lnx+c3x-2(lnx)2 16 y = c1x + c2x2 + 4

15xndash1 21 2

17 y = 2x4+x-3 18 y = 3117

x + 317

xndash 2cos 5lnx ndash 7685

xndash 2sin 5lnx

19 y = 53

x2 ndash 23

x5 20 y = 13

10x3 + 13

15xndash 2 ndash x

6

21 (i) y = (1ndashln (x+1)) (x+1) (ii) y = [1ndash4ln(1ndashx)](1ndashx)3

(iii) y = x ndash 21 21 2 3cos 3

2ln x ndash 2 + 5

3sin 3

2ln x ndash 2

55B Προβλήματα

1 y x = c1

ndash 1 n 4n

2n + 1 Σ

n = 0

infinxn + c2 x ndash 1 21 2 ndash 1 n 4n

2n Σ

n = 0

infinxn

2 y x = c1 xndash 1 21 2 sin x + c2 xndash 1 21 2 cos x

3 y x = c1 x1 41 4 1 +ndash 1 n

n 59 4n + 1xnΣ

n = 1

infin+ c2 1 +

ndash 1 n xn

n 37 4n ndash 1Σn = 1

infin

4 y x = c1 x 2 1 + 1n 1 + 2 2 2 + 2 2 n + 2 2

xnΣn = 1

infin

+ c2 xndash 2 1 + 1

n 1 ndash 2 2 2 ndash 2 2 n ndash 2 2xnΣ

n = 1

infin

5 y x = c1x3 1 +n + 1 n + 2

1013 3n + 7xnΣ

n = 1

infin+ c2 x2 32 3 1 +

3n ndash 4 3n ndash1

n3n xnΣn = 1

infin

6 y x = c1 x 5 1 +1 + 5 2 + 5 n + 5

n 1 + 2 5 2 + 2 5 n + 2 5xnΣ

n = 1

infin

+ c2 xndash 5 1 +

1 ndash 5 2 ndash 5 n ndash 5

n 1 ndash 2 5 2 ndash 2 5 n ndash 2 5xnΣ

n = 1

infin

7 y x = c1 x1 31 3 + 15

x4 34 3 + c2 xndash 1 31 3 ndash 1 n 10

3n n 3n ndash 2 3n ndash 5Σ

n = 0

infinxn

8 y x = c1 xndash 1 41 4 an ndash 14Σ

n = 0

infinxn + c2 xndash1 anΣ

n = 0

infinndash1 xn

όπου an λ = ndash2 n + λ ndash1

n + λ 4n + 4λ + 5 + 1an ndash 1

9 y x = c1ndash 1 n

n 1 + 2 2 n + 2 2Σ

n = 0

infinxn + 2

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 24: 978-960-7182-07-4

26 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2

ndash 1 n

n 1 ndash 2 2 n ndash 2 2Σ

n = 0

infinx n ndash 2

10 y x = C1 y1 + C2 y1 όπου

y1 x = cos ln x + i sin ln x 1 +xn

n 1 +2i 2 + 2i n + 2iΣ

n = 1

infin

11 y x = c1 1 +2n x + 2 n

n 57 2n + 3Σn = 1

infin

+ c2 x + 2 ndash 3 23 2 1 +

2n x + 2 n

n ndash 1 13 2n ndash 3Σn = 1

infin ndash infin lt x lt ndash2

ndash2 lt x lt + infin

12 y x = c1 Σn = 0

infin 1n 2n + 1

xn + c2 xndash1 21 2 Σn = 0

infin 1n 2n ndash 1

xn

13 y x = c1 x3 23 2 1 + 3xn

n 2n + 3 Σn = 1

infin+ c2 1 ndash x ndash

xn

n 2n ndash 3 Σn = 2

infin

14 y x = c1 x1 31 3 ndash 1 n 2nxn

n 47 3n + 1Σn = 0

infin+ c2

ndash 1 n 2nxn

n 25 3n ndash1Σn = 0

infin

15 y x = c1 x 1 +xn

n 711 4n + 3Σn = 1

infin+ c2 xndash 1 21 2 x2n

n 15 4n + 1Σn = 0

infin

16 y x = c1 x1 21 2ndash 1 n xn

n 2nΣn = 0

infinequiv x1 21 2 endash x 2x 2 + c2 1 +

ndash 1 n xn

2n ndash1 Σn = 1

infin

17 y x = c1 x1 21 2 1n 2nΣ

n = 0

infinx2n + c2

sin 3xx

18 y x = c1 cos x + c2 sin x x gt 0

19 y x = c1cos 3x

x + c2sin 3x

x x ne0 20 y x = c1cos h 2x

x +c2sin h 2x

x

21 y x = c1cos x 2x 2

x + c2sin x 2x 2

x 22 y x = c1 cos x2 + c2 sin x2

23 y x = c1 x1 21 2 cos h x + c2 x1 21 2 sin h x

24 y x = c1 ndash c2 endash x + c2 xndash1 ex 0 lt x lt infin

25 y x = c1 x ex + c2 e2x 26 y x = c1 1 ndash x ndash 1 ndash c2 xndash 1 21 2 1 ndash x ndash1

27 y x = c1 xndash1 cos 2x + c2 xndash1 sin 2x

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 25: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 27

28 y x = c1 1 + x5

ndash 3100

x2 + cos ln x + x25

10 + x + sin ln x

+ c2 1 + x

5ndash

3 x2

100+ sin ln x ndash x

2510 + x + cos ln x

29 y1 x = x y2 x = x ln x + 1n n

xnΣn = 1

infin

30 N = 0 rArry x = 1 N = 1 rArr y x = xndash1 1 ndash x

N = 2 rArr y x = xndash2 1 ndash 2x + 12

x2 N = 3rArry x = xndash1 1 ndash 3x + 32

x2 ndash 16

x3

31 y x = c1 F 2 ndash 1 32

x + c2 xndash1 2 F 32

ndash 32

12

x

= c1 1 ndash 4

3x + c2 xndash 1 21 2 F 3

2 ndash 3

2 1

2 x

32 y = c1 F 12

1 12

ndash x + c2 ndash x12 F 1 3

2 3

2 ndash x = c1

11 + x

+ c2ndash x 1 21 2

1 + x

33 y x = c1 F 2 2 12

x + 12

+ c2x + 1

2

1 21 2F 5

2 5

2 3

2 x + 1

2

34 y x = c1 F 1 1 145

3 ndash x5

+ c23 ndash x

5ndash 9

5 F ndash 45

ndash 45

ndash 45

3 ndash x5

35 y x = c1 F 1 ndash 1 ndash 1 21 2 1 ndash ex + c2 1 ndash ex32 F 5

2 1

2 5

2 1 ndash ex

55Γ Προβλήματα

1 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x με ϕ1 x =ndash 1 n xn

n n + 1 Σ

n = 0

infin

ϕ2 x = ndash ϕ1 x ln x + x ndash1 1 +

ϕ n + ϕ n ndash 1

n n ndash 1 ndash 1 n + 1 xnΣ

n = 1

infin

όπου ϕ n = A J1 n + B Y1 n Α Β αυθαίρετα |x|gt0

2 λ1 = λ2 = 1

y x = c1 x 1 + x + c2 x 1 + x ln x ndash 2 x 2 1 +ndash1 n + 1

2n n + 1xnΣ

n = 1

infin

3 λ1 = 0 λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n

nΣn = 0

infinxn + 1

ϕ2 x = ndash ϕ1 x ln x + b nΣ

n = 0

infinxn

με b0 = 1 b1 = 0 και n b n + 1 + b n =ndash 1 n

n n = 1 2 x gt 0

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 26: 978-960-7182-07-4

28 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

4 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =4n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

22n + 1

n 2Σn = 1

infinHn xn

5 λ1 = λ2 = 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =ndash 1 n xn + 1

nΣn = 0

infin ϕ2 x = ϕ1 x ln x ndash

ndash 1 n

nΣn = 1

infinHn xn + 1 x gt 0

6 λ1 = λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = xndash1 + 3 ϕ2 x = ϕ1 x ln x ndash 9 +

ndash 3 n xn ndash 1

n n ndash 1 nΣn = 2

infin x gt 0

7 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x =3n xn

n 2Σn = 0

infin ϕ2 x = ϕ1 x ln x ndash

2 3n

n 2Σn = 1

infinHn xn x gt 0

8 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = 2 + x ϕ2 x = ϕ1 x ln x + xndash1 ndash 3 ndash 4x +

2 ndash 1 n xn ndash 1

n n ndash 1 n ndash 2Σn = 3

infin

9 λ1 = 0 λ2 = 2 y x = c1 ϕ1 x + c2 ϕ2 x

όπου ϕ1 x = x2 και ϕ2 x = ϕ1 x ln x ndash 1 + 2x + 2

ndash x n

n n ndash 2 x gt 0 Σ

n = 3

infin

10 λ1 = 0 λ2 = ndash 1 y x = c1 ϕ1 x + c2 ϕ2 x

ϕ1 x = xndash1 12n Σ

n = 0

infinx2n = 1

x cos h x ϕ2 x = xndash1 12n + 1 Σ

n = 0

infinx2n + 1 = 1

x sin h x

11 λ1 = 4 λ2 = 0 y x = c1 1 + 23

x + 13

x2 + c2 n + 1Σn = 0

infinxn + 4

12 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x ndash x + 1

4x2 ndash 1

3 3x3 + 1

4 4x4 όπου ϕ1 x = 1

nΣn = 0

infinxn = ex

13 λ1 = λ2 = 0 y x = c1 ϕ1 x + c2 ϕ1 x ln x

+ ϕ1 x 2x + 5

4x2 + 23

27x3 + όπου ϕ1 x =

ndash 1 n

n 2nΣn = 0

infinxn

14 λ1 = λ2 = 1 y x = c1x endashx + c2 x endashx ln x + x + 14

x2 + 13 3

x3 +

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
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  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 27: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 29

15 λ1 = 2 λ2 = 0 y x = c1 x2 + c212

x2 ln x ndash 12

+ x ndash 13

x3 +

16 λ1 = 7 λ2 = 0

y x = c1 1 ndash x

2+

x2

10ndash

x3

120+ c2 x7 +

ndash1 n 4 5 6 n + 3

n 8 9 10 n + 7xn + 7Σ

n = 1

infin

17 λ1 = 2 λ2 = ndash 2

y x = c1

xn + 2

n n + 4 Σn = 0

infin+ c2

xn + 2

n n + 4 Σn = 0

infinndash 6 xndash2 1 ndash x

3+

x2

12ndash

x3

36+

18 λ1 = λ2 = ndash 1

y x = c1 xndash1 1 ndash x + c2 xndash1 1 ndash x ln x + xndash1 3x ndashx2

4ndash

n ndash 2

n 2xnΣ

n = 3

infin

19 λ1 = 32

λ2 = ndash 32

y x = c1 x3 23 2 1

n + 3 Σn = 0

infinxn + c2 xndash 3 23 2 1 + x + 1

2x2

= c1 xndash 3 23 2 ex ndash 1 ndash x ndash

x2

2+ c2 xndash 3 23 2 1 + x +

x2

2

20 λ1 = λ2 = 1 y x = c1 x + c2 x ln x +ndash 1 n

n nxn + 1Σ

n = 1

infin

21 (ii) k = n ακέραιος (iii) L0 (x) = 1 L1 (x) = 1 ndash x L 2 x = 1 ndash 2x +x2

2

(iv) y1 = 1 y2 = ln x + x +x2

2 2+

x3

3 3+

x4

4 4+

(v) y1 = 1 ndash x y2 = 1 ndash x ln x + 3x ndashx2

4ndash

x3

36ndash

x4

288ndash

22 (ii) k ne ndash 14

+m2

4 misinΖZ ϕ1 x = xλ

1 an xnΣn = 0

infin ϕ2 x = xλ

2 an xnΣn = 0

infin

(iii) k = ndash 14

ϕ1 x = x1 21 2 an xnΣn = 0

infin ϕ2 x = ϕ1 x ln x + x1 21 2 b nΣ

n = 1

infinxn

(iv) k = ndash 14

+m2

4 m ne 0

29 y x = c1 J3 3x + c2 Y3 3x

30 y x =

c1 Jn x 7 + c2 Yn x 7 για v = n isinZΖ

c1 Jv x 7 + c2 Yndash v x 7 για v notinΖΖ

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 28: 978-960-7182-07-4

30 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

31 y x = c1 J3 x 6 dx + c2 Y3 x 6 dx + c3

32 y x = c1 J1 31 3x + c2 Jndash 1 31 3

x 33 y x = c1 J5 25 2x + c2 Jndash 5 25 2

x

34 y x = c1 J2 3x + c2 Y2 3x 35 y x = c1 x4 J1 x + c2 x4 Y1 x

36 y x = c1 xndash 1 J2x2

+ c2 xndash1 Y2x2

37 y x = c1 x2 J4 x2 + c2 x2 Y4 x2

41 y1 x = x1 21 2 J1 21 2x y2 x = x1 21 2 Jndash 2 1 2

x

42 y1 x = xndash1 Jndash1 x y2 x = xndash1 J1 x = ndash xndash1 Jndash 1 x

43 y1 x = x J3 23 2x y2 x = x Jndash 3 23 2

x

56 Προβλήματα

1 ΚΑ 2 ΜΚΑ

3 ΚΑ 4 ΜΚΑ

5 ΚΑ 6 ΚΑ

7 ΚΑ 8 ΚΑ

9 ΜΚΑ 10 ΜΚΑ

11 ΜΚΑ

12 y x = c1 1 + 2x + 7

3 x2+ 112

45 x3+

+ c2 xndash 1 21 2 1 + 4

3x+ 22

15 x2

+ 484

315 x3

+ x gt 1

13 y x = c1 1 ndash 13x

+ 130 x2

ndash 1630 x2

+

+ c2 x

1 21 2 1 ndash 1x + 1

6 x2

ndash 1

90 x2

+ x gt 0

14 y x = c1 xa 1 ndash Σn = 1

infin ndash 1 n a a ndash 1 a ndash 2n + 1

2n n 2a ndash 1 2a ndash 2n + 1xndash2n

+ c2 xndash a + 1 1 +

a + 1 a + 2n xndash2 n

2n n 2a + 3 2a + 2n + 1Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars 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HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 29: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 31

15 y x = c1xndash 2n

2n n 2n ndash 1 2n ndash 3 2n ndash 5Σ

n = 0

infin+ c2 xndash 1 ndash 2

3xndash 3 + 1

15xndash 5

16 y x = c1xndash 3 23 2 1 +ndash 2 n n + 1 xndash n

5 7 9 2n + 3Σ

n = 1

infin+ c2

ndash 1 n + 1 2n ndash 1 xndash n

nΣn = 0

infin

17 y x = c1 x2 + 2x + 3 + c2 xndash1 n + 4Σn = 0

infinxndash n

18 y x = c1 xndash 1 + c2 xndash 1 21 2 xndashn

2n ndash 1Σn = 0

infin

19 y x = c1 xndash1 n + 1 2n + 3 2n + 5 xndash nΣn = 0

infin

+ c2 xndash 1 21 2 n + 1 n + 2 2n + 1 xndash nΣ

n = 0

infin

20 y x = c1 xndash1 ndash 1 n + 1Σn = 2

infinn n ndash 1 xndash n +

+ c2 xndash1 ln xndash1 ndash 1 n + 1 n n ndash1 xndash n + xndash 1 + xndash 2Σ

n = 2

infin

+xndash1 ndash 1 n + 2 n2 + n ndash 1 xndash nΣ

n = 2

infin

21 y x = c1 xndash1 n xndashnΣn = 1

infin+ c2 xndash1 ln xndash1 n xndashn + xndashnΣ

n = 0

infinΣ

n = 1

infin

22 y x = c1 cos 2 xndash 1 + c2 sin 2 xndash 1

57 ΠΡΟΒΛΗΜΑΤΑ ΕΠΙΣΚΟΠΗΣΗΣ

4 (i) όχι (ii) ϕ2 x = x ex2

(iii) ϕ1 x = 83

ndash 5x3

ex2ndash1 ϕ2 x = 8 ndash 5x ex2ndash1

9 λ1 = λ2 = 0

y1 x = 1 +ndash 1 n + 1 a a + 1 1 middot 2 ndash a a + 1 n n ndash 1 ndash a a + 1

2n n 2Σn = 1

infinx ndash 1 n

= F ndash a a + 1 1 1 ndash x

2

10 λ1 = 12

λ2 = 0 y x = c1 x ndash 11 21 2 1 +

ndash 1 n 1 + 2α 2n ndash 1 + 2α 1 ndash 2α 2n ndash 1 ndash 2α2n n + 1

Σn = 1

infinx ndash 1 n +

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 30: 978-960-7182-07-4

32 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

+ c2 1 +ndash 1 n α 1 + α n ndash 1 + α ndash α 1 ndash α n ndash 1 ndash α

n 1 3 5 2n ndash 1Σn = 1

infinx ndash 1 n

11 y x = c1 F ρ σ τ x ndash α1

α2 ndash α1

+ c2

x ndash α1α2 ndash α1

1 ndash τF ρ ndash τ + 1 σ ndash τ + 1 2 ndash τ

x ndash α1α2 ndash α1

όπου ρ + σ + 1 = ba τ =

b α1 ndash α3

a α1 ndash α2

ρ σ= ca

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 31: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 6

61 Προβλήματα

10 xgt0 11 12 xlt0

13 xgt0 14 ndash π2

lt x lt π2

15 ndash infin lt x lt infin

62 Προβλήματα

1

x = ndash c1 e3t + c2 endash t 2

y = c1 e3t + c2 endash t

2

x = c1 + c2 endash t+ 1 21 2 et+ 5t3

y = c1 ndash 2c2 endash t+ 5t3

3

x = c1 et+ 1 41 4 cost ndash 1 41 4 sint

y = ndash 3 c1 et ndash 3 43 4 cost ndash 1 41 4 sint 4

x = sint ndash costy = sint

5

x = c3 e2t+ 1 21 2 et c1 ndash c2 cost + c1 + c2 sint

y = et c1 cost + c2 sint

z = 3 23 2 et c1 ndash c2 cost + c1 + c2 sint + c3 e2t

6 x = c1 cos 2 t + c2 sin 2 t + 3 23 2 t

y = ndash 2 c1 sin 2 t + 2 c2 cos 2 t + (32) ndash t2

7 x = c1 e3t+ c2 cos 2t + c3 sin 2t

y = ndash 5c1 e3t + c3 cos 2t ndash c2 sin 2t ndash1

8

x = sinty = c1t + c2

9 x = sinty = 0

10 όχι λύση

11

x = 4c1 e3t ndash 3c2 endash 4t 6

y = c1 e3t + c2 endash 4t

z = ndash 4c1 e3t + 3c2 endash 4t 6

12

x = c1 + c2 e4t+ c3 e8t

y = 2c1 ndash 2c3 e8t

z = 2c1 ndash 2c2 e4t + 2c3 e8t

13

x = c1 e2t + c2 endash t + 2 32 3 tendash t

y = c1 e2t + c3 endash t ndash 1 31 3 tendash t

z = c1 e2t ndash c2 + c3 + 1 31 3 endash t ndash 1 31 3 tendash t

14 x = 4 e3tndash endash 2t 5

y = 2 6e3t ndash e ndash 2t 5

15 x = 3e2t ndash e ndash 2t 2

y = 3e2tndash 5endash 2t 2 16

x = endash 2t 3 cos 3t + 9 sin 3t

y = endash 2 t 2 cos 3t ndash 4 sin 3t

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
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Page 32: 978-960-7182-07-4

34 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

17

x = et

y = ndash et

18

x = ndash 5e2tsint

y = ndash e2t 3 sint + cost

19

x = et2et2 ndash e3t

4e3t4 ndash endash

4e t4

y = et ndash e3t4e3t4 ndash 3endash t

43e 4

20 x = c1 e 1 + 6 t+ c2 e 1ndash 6 t+ 11 sint ndash 7 cost 20

y = c1 1 + 6 3 e 1 + 6 t + c2 1 ndash 6 3 e 1 ndash 6 t + ndash cos t + 3 sin t 20

όπου

c1 = 33 ndash 6 33 4033 32 3 c2 = 33 + 32 6 36 3 402 6 3

21 όχι λύση 22 άπειρες λύσεις Ορίστε αυθαίρε-

τα το x(t) και λύστε ως προς y(t)

23 άπειρες λύσεις x t = ndashy t 24 όχι λύση

25 άπειρες λύσεις 26 όχι λύση

27 άπειρες λύσεις 28 όχι λύση

29

x = g t y = 1 21 2 t2Πg t + g t dt

g t αυθαίρετη παραγωγίσιμη συνάρτηση του t

30 όχι λύσεις 31 άπειρες λύσεις που ικανοποιούν

τη σχέση x+y=et+endash2t

63 Προβλήματα

1 ναι 2 ναι 3 ναι 4 όχι

5 ναι 6 όχι 7 ναι 8 ναι

9 ναι 10 ναι 11 ναι 12 ναι

13 x(t) = 7x1(t) + 3x2(t) + 5x3(t) 14 x(t) = 3x1(t) ndash 3x2(t) ndash 5x3(t)

15 x(t) = 3x1(t) ndash 2x2(t) 16 x(t) = x1(t) + 2x2(t) + x3(t)

17 x (t) =ndash [x1(t) + x3(t)] (718)

18 (i) x(t)=3x1(t)+7x2(t)+x3(t)ndash2x4(t) (ii) x(t)=13 x1(t)+41 x2(t)+3x3(t)ndash 12x4(t)

64 Α Ι) Προβλήματα

1 x = c112

endash t+ c2ndash 2

1endash 4t 2 x t = c1

111

e2t+ c2

10

ndash 1endash t+ c3

01

ndash 1endash t

3 x t = c1

111

e4t+ c2

1ndash 21

et+ c3

10

ndash 1endash t

4

x t = c1

1i

+ c21ndash i

e2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 33: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 35

5 x t = c1

101

e3t+ c2

ndash 110

e3t+ c3

ndash 1ndash 11

endash 3t

6 x t = c1

1ndash 11

e5t+ c2

110

endash t+ c3

ndash 101

endash t

7 x t = c1

1ndash 10

endash 3t+ c2

111

e3t+ c3

11

ndash 2e3t

64 A II) Προβλήματα

1 e

At= 1 + 2t ndash t

4t 1 ndash 1te

t 2 x t = c1

11

endash t 2t 2 + c2

11

te

3

eAt= e2t1 t t2

2t22

0 1 t0 0 1

4

eAt= 15

endash 2t+ 5e2tndash e3t 5e2tndash 5e3t endash 2 tndash e3t

ndash e ndash 2 t+ e3t 5e3t ndash endash 2t+ e3t

4endash 2 tndash 5e2t+ e3t ndash5e2t+ 5e3t 4endash 2t+ e3t

5

eAt

= 15

6e2t

ndash cost ndash 2sint ndash2e2t

+ 2cost + 4sint 2e2t

ndash 2cost + sint

3e2t

ndash 3cost ndash sint ndashe2t

+ 6cost + 2sint e2t

ndash cost + 3sint5sint ndash 10sint 5 cost

6

eAt =

t + 1 endash t t + 1 endash t ndash endash 2t endash tndash endash 2t

ndash t endash t ndash t endash t+ endash 2 t endash 2 tndash endash t

t endash t t endash t endash t

7 A = 1 00 2

8 A =1 0 00 1 00 0 3

9 A =2 0 00 2 00 0 2

10 eAt

= e2t 1 ndash t ndasht

t 1+ t 11 eAt = endash t

1 + 3t ndash 3 23 2 t2

ndash 3

9t ndash 9 29 2 t2

t ndash t + 1 21 2 t2

t 1 t

3t 1 ndash 3t + 3 23 2 t2

12 eAt

= endash 2 t

1 0 04t 1 0

t 0 1 22

eAt =1 0 0t 1 0

t22 t 1

24 A =3 1 ndash 11 3 ndash 13 3 ndash 1

25 A = 113

16 ndash 25 308 ndash6 ndash240 13 26

26 όχι

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 34: 978-960-7182-07-4

36 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

64 Α III) (a) Προβλήματα

1 x t = c1

13

e4t+ c21

ndash 3endash 2 t

2

x t = c1

1ndash 4

endash 2t+ c211

e3t

3 x t = c111

et+ c253

endash 3t

4

x t = c1

1ndash 1

endash t+ c211

e3t

5 x t = c1

1ndash 1

endash t + c232

e4t

6 x t = c1

1ndash 6

endash 2 t+ c211

e5t

7

x t = c1

111

e9t+ c2

1ndash 21

e6t + c3

10

ndash 1 8 x t = c1

111

e6t+ c2

1ndash 21

e3t + c3

10

ndash 1e3t

9 x t = c1

625

+ c2

312

et+ c3

212

endasht

10

x t = c1

1ndash 10

e2t + c2

01ndash 1

endash2t+ c3

1ndash11

e3t

11 x t = c1

ndash 211

endasht+ c2

0ndash11

endash3t+ c3

111

e5t

12

x t = c1

ndash101

endasht+ c2

1ndash13

endash2t + c3

143

e3t

13 x t = 11

e3t

14 x t = 1

3ndash 21

endash 4 tndash 13

11

e5t

15 x t = 22

e3tndash 128

e4t

16 x t =

5 65 615 615 6

e4t+1 61 6

ndash1 21 2endash 2 t

17 x t =11ndash1

e2t +ndash 12ndash 212

endasht +121

12endash3t

18 x t = ndash 5

6

100

etndash 12

ndash 5ndash 22

endash t+ 13

ndash 203

endash 2t

19 x t = 38

ndash 2ndash 11

ndash 18

2ndash 33

e8t

20 x1 t = x4 t = 2 e10 t

+ e15t

x2 t = x3 t = ndashe10 t

+ 2e15t

22 i x1=

λ2ndash λ1λ1λ2

endashλ1t x2=01

ndash 1endashλ2t x3=

001

24 x t = c13t2

t2+ c2

t4

t4

25 x t = c111

t + c213

tndash 1

26

x t = c1

12

tndash 1+ c221

t2

27 x t = c134

+ c212

tndash 2

28 x t = c1

32

t4+ c21

ndash 1tndash 1

29 x t = c111

+ c2

V2ndash V1

ek 1 V11 V1

+ 1 V21 V2

t

30

P1 t

P2 t= c1

11

endash 2 t+ c252

et

64 A III) (b) Προβλήματα

1 x t = endash 2t 2 2 sint1 ndash sint cost + sint

2 x t = et2 0 0

ndash 2 cos 2t sin 2t3 sin 2t ndash cos 2t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars 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HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 35: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 37

3 x t = cos πt sin πt

ndash sin πt cos πtendash 2t

4

x t = cos 2 t sin 2 t

ndashsin 2 t cos 2 tendash 3t

5

x t =3e3t et cos 2t et sin 2t4e3t et cos 2t + sin 2t et sin 2t ndash cos 2t

4e3t 0 0

6 x t =2 0 0

ndash 3 cos 2t sin 2t2 sin 2t ndash cos 2t

et

7 x t = endash 2t cost ndash 5 sint

ndash 2 cost ndash 3 sint

8 x t = e3t 2cost + sintndash cost ndash 3 sint

9 x t =et

2

1 ndash 3cos 2t + 2sin 2t3sin 2t + 2cos 2t

3 ndash 3cos 2t + 2sin 2t

10 x t =etndash10

2

1 ndash cos 2 t ndash 10sin 2 t ndash 10

3 ndash cos 2 t ndash 10

11 x t =2

ndash 21

endash 2t

+ndash 2 sin 2 t ndash 2 cos 2 t

cos 2 t ndash 2 sin 2 tndash 3 cos 2 t

endash t

12 x t = ndash25ndash 76

et ndashcos 5t ndash 5 sin 5t

cos 5tcos 5t

+ 65cos 5t + sin 5t

sin 5tsin 5t

13 (ii) Ψ t =

ndash 2 6 sin 6 t 2 6 cos 6 t sin t ndash cos tndash 12 cos 6 t ndash12 sin 6 t cos t sin t

6 sin 6 t ndash 6 cos 6 t 2sin t ndash2cost6 cos 6 t 6 sin 6 t 2cost 2sin t

(iii) x t = 2

52y0 ndash z 0 cos 6 t + 1

5y0 + 2z0 cost ndash 12

5 62y0 ndash z 0 sin 6 t

ndash 15

y0 + 2z0 sint ndash 15

2y0 ndash z 0 cos 6 t + 25

y0 + 2z0 cost

65

2y0 ndash z 0 sin 6 t ndash 25

y0 + 2z0 sintT

14 x(0) = (x1 0 x3)T 15

x t = ctndash 1 cos 2 ln t

2 sin 2 ln t+ c2tndash 1 sin 2 ln t

ndash 2 cos 2 ln t

16 x t = c1

5cos ln t2 cos ln t + sin ln t

+ c25 sin ln t

cos ln t + 2sin ln t

17 x t = c1

10

ndash 2tndash 1 + c2

cos ln tsin ln tcos ln t

tndash 1 + c3

sin ln tsin ln tcos ln t

tndash 1

64 A III (c) Προβλήματα

1

x t = c112

+ c212

t ndash 01 21 2

2 x t = c111

endash t 2t 2 + c2

11

tendash t 2t 2 + 0

2 52 5e

ndash t 2t 2

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 36: 978-960-7182-07-4

38 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

3 x t = c1

13

+ c213

t +1 41 4

ndash1 41 4 4 x t = c1

11

et+ c2

11

te2t

+ndash1 31 3

0e

2t

5 x t = et 2 2t ndash 1

1 t 6

x t = e2t

1 0 0t 1 0

ndash t ndash 1 1

7

x t = c1

111

et+ c2

110

e2t

+ c3

101

e2t

8 x t = c1

ndash 4ndash 5

2+ c2

20

ndash 1e

5t+ c3

20

ndash 1te

5tndash

112

e5t

2

9 x t =

endash 3t

endash 3t

e3t

ndash endash 3t

endash 3 t

e3t

0 endash 3t

ndash 2e3t

10 x t = e2t1 t t2

2 1 + 2t 2t2+ 2t

4 4 + 4t 4t2+ 8t + 4

11 x t = c1

010

endash 2t

+ c2

110

+ c3

tt1

endash t

12 x t = c1

001

endash 2t

+ c2

100

+ c3

+tndash10

endash t

13 x t =101

et

14 x t =

1 ndash ttt

e2t

15

x t = e

3t 1 + 2t1 ndash 2t

16 x t = e3t 4t

4t ndash 1 17 x t = e

2t 010

18

x t =ndash 12

ndash 33e

t+ 4

01

ndash 6te

t+ 3

001

e2t

19 x t = 3 + 4t

2 + 4te

ndash 3t

64 B Προβλήματα

1 x t = c121

et 2t 2+ c2103

e3t 23t 2 ndash13 213 213 413 4

tet 2t 2 ndash15 215 29 49 4

et 2t 2

2 x t = c121

et+ c2

11

e2t

+ 3 11

et+ 2 2

1te

t

3 x t = c11

ndash 1e

t+ c2

ndash t1 21 2 ndash t

et+

1 21 2ndash 2

endash t

4 x t = c1cos tsin t

+ c2sin t

ndash cos t + cos tsin t

t + ndash sin tcos t ln cos t

5 x t = c1

cos tsin t

et+ c2

sin tndash cos t e

t+ cos t

sin tte

t

6

x t = c1

111

e2t

+ c2

ndash 101

endash t

+ c3

ndash 110

endash t

+1

ndash 1ndash 1

et

7 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2cos t + sin t

+ c3

ndash sin tcos t + sin t 2

sin t ndash cos t+

2ndash 34

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 37: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 39

8 x t = c1

010

et+ c2

ndash cos tcos t ndash sin t 2

cos t + sin t+ c3

ndash sin tcos t + sin t 2cos t + sin t 2sin t ndash cos t

+ndash 1t0

et

9

x t = c1

ndash 101

+ c2

ndash 120

endash t

+ c3

ndash102

endash t

+2t ndash 1 ndash 2e

ndash tndash 2 te

ndash t

02t +2 + 2e

ndash t+ 4te

ndash t

10

x t = c1

2ndash 32

et+ c2

0cos 2tsin 2t

et+ c3

0sin 2t

ndash cos 2tet+

ect c ndash 1 ndash1

1 ndash c 2 + 4

4 + 1 ndash c 2

2 c ndash 41 + 3c

11 x t = 2et t cost + 3t sint + sint

ndash tsint 12 x t = ndash cos t ndash sin t + 1

sin t ndash cos t + 1

13 x t = ndash 11

et+ 1

2e

2t+ 0

ndash 1e

3t

14 x t = 1 + t

22t

22

t e2t

15 x t =27 2527 25 e2tndash 3 53 5 te2tndash 1 251 25 14 sint + 2 cost

1 51 5 sint ndash 2 cost ndash 3 53 5 e2t

16

x t =

etndash e

ndash t

endash t

1 + t

0

17 x t =

3e3t

ndash 2e2t

ndash te2t

e2t

3e3t

ndash 2e2t

19

xp t =e3t 1

2e3t ndash 1

3 43 41

+endash t

4e 4ndash endash 2 t ndash endash t

2t

2

20 x t =1

ndash 1ndash 1

te3t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 38: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 7

72 Προβλήματα

1 ΤΣ 2 Σ 3 Α

4 ΤΣ 5 Σ 6 Α

7 c=2 8 όχι 9 cgt1

10 c=1 11 c=1 12 c=1

13 c=1 14 c=0 15 c=1

16 c=1 17 c=0 18 c=a

19 όχι 20 c=2 21 c=1

73 Προβλήματα

1 F s =3s2 + 2s + 3

s2 + 1 2

2 F s = π 2s3 23 2 ndash1+ 3 π 2s5 25 2

3 F s = as2 ndash a2

4 F s = ln s

s2 + a2

5 F s = ln s ndash bs ndash a 6

F s = s2 + 2a2 s s2 + 4a2 ndash1

7 F s = 6a3 s2 + a2 s2 + 9a2 ndash1 8 F s = 2a2 s s2 + a2 s2 + 9a2 ndash1

9 F s = 2 s ndash a ndash 3 10

F s = 2s s2 ndash 3b2 s2 + b2 ndash 3

11 F s = 6 s ndash a ndash 4

12 F s = s2 + a2 s2 ndash a2 ndash 2

13 F s =2 sa s2 ndash a2 ndash 2 14 F s =s3 s4 + 4 a4 ndash1

15 F s = a s2 + 2a2 s4 + 4a4 ndash1 16 F s = a s2 ndash 2a2 s4 + 4a4 ndash1

17 F s = 2a2 s s4 + 4a4 ndash1

18 F s = nπ s + 2 2 + n2 π2 ndash1

19 F s = A s + a + Bβ s + a 2 + β2 ndash1

20 F s = 1

s ndash 4s2

+ 4

s + 3 3

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
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Page 39: 978-960-7182-07-4

42 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

21 F s = 6s

s2 + 92

ndash 1s + 4

s3

22 F s = 2ndash1 s ndash 6 ndash1 + s + 6 ndash1 + 2sndash1

23 F s = 3 s ndash 3 ndash 2ndash 3 s + 3 ndash 2ndash 54 sndash 4

24 F s = 4 s + 1 ndash 3ndash 2s ndash1 + s 2 s2 + 4 ndash 1

25 F s = s 2 s2 ndash 16 ndash 1ndash s 2 s2 + 16 ndash1

26 F s = s2

ndash 9 s2

+ 9ndash 2

ndash s s2

+ 9ndash 1

+ 5 s + 1ndash1

27 F s = s ndash 3 ndash1 ndash 12 sndash 4 28 F s = sndash1 ndash 6 s2 ndash 36ndash1

+ s + 1 s + 1 2 ndash 1ndash1

29 F s = 6 s + 4 2+ 4ndash1ndash s s2 + 36 ndash1 + 6sndash1

30 F s = 2 s ndash3 ndash1 1 ndash endash sndash 3

1 + endash sndash 3

31 F s = h

s1 ndash endash 4s

1 + endash 4s 32 F s = 1 + endash πs 1 ndash endash2 πs s2 + 1 ndash1

33 F s = 1 ndash endashs 2 s 1 ndash endash 2s ndash1 34 F s = 1 + endash πs 1 ndash endash πs s2 + 1 ndash1

35 F s =

e1 ndash s ndash11 ndash s 1 ndash endashs

36 F s = 1

1 ndash endash πs2 ndash endash πs 2πs 2 πs + 2

π s2+

sendash πs 2πs 2 + endash πs

s2 + 1

37 F s = 1

as2tanh as

2 38

F s = a

s1bs

ndash 1ebs ndash 1

39 G s = a s ebt ndash 1ndash1

40

(a) 1 ndash 2s2 ndash1

+ 1 3 1middot2 s4ndash1

ndash = 1 + s2 ndash1 21 2

(c) L e r f t = 2

πndash 1 k 2 k

k s2 k + 2Σk = 0

infin

(d)

s s + 1 1 21 2 ndash1

42 (i) F 3i = ndash 2

5= F ndash 3i

(ii)

F a + ib = 2

a + ib 2 + 4= F a ndash ib

74 Προβλήματα

1 1

33 23 2sin 3 t ndash 3 t cos 3 t ndash 1

180t6

2 1

6e3t ndash 1

6endash 3t ndash 7 cos 15 t

3

t3

6ndash

t4

12+

t5

30 4

23

e3t + 13

endash 3t ndash1

5 cos 6 t ndash 5

6sin 6 t

6

ndash 3

2t2 endash 2t + 4 cos 6 t

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 40: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 43

7 43

sin 3t ndash t e3t

8 43

e2t ndash endash t

9 et ndash 2t et

10 73

endash 2t ndash 13

et

11 t2 et 1 ndash 1

3t

12 e2t cos 15 t

13

23

et + 13

cos 2 t ndash 13 2

sin 2t

14 ndash 7209

882e2t ndash 1890

441t e2t + 12299

882e4t + 983

441endash5t

15

sin 2 n π t

T

16

t3 23 2

Γ 1 21 2 17 1

2 ω3sin ωt ndash ωt cos ωt

18

12ω

sin ωt + ωt cos ωt

19 endash t sin t

t

20 e

t ndash1t

21

endash bt ndash endash at

t

22 4t sin 2t + 3 sin 2t ndash 6t cos 2t16

23 t2 endash 3t 1 + t

6

24 7 cos 3t + 4 sin 3t

27 L f1 t = L f2 t = L f3 t = sndash 2 Lndash1sndash2 = f3 t

28 L f1 t = L f2 t = L f3 t = s ndash 1 ndash1 Lndash1 s ndash 1 ndash1 = f3 t

30 L ndash1

F s = 43

ndashendash t

8ndash 7

4et + 13

14e3t

31

L ndash1

F s =endash t

4ndash 5

4e3t

32 L ndash1

F s = ndash 23

+3 et

4ndash

endash 3t

12 33

L ndash1

F s =5 endash t

2ndash 9 endash 2t +

15 endash 3t

2

34

320

et ndash 14

endash t +cos t endash 2t

10ndash

sin t endash 2t

5

36 f t = 2 endash t ndash 4 e3t + 5 e2t 37 f t = 1 ndash 3 endash t + 3 endash 2t

38 f t =

3 ndash 10t et

50+

endash t ndash 9 cos 3t + 13 sin 3t150

39 f t = 1

253 cos t + 4 sin t ndash 3 endash 2t ndash 10 t endash 2t

40 f t = 4

25+ 2t

5+

2 t2

5+

t3

6endash 2t

54 + ndash 425

+ 2t5

ndash2 t2

5+

t3

6e3t

54

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 41: 978-960-7182-07-4

44 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

75 Προβλήματα

1 N = ndash 1

144+ t

12+ endash t2 cos 47 t

2ndash 23

47sin 47 t

2 144

2 y = ndash t 7t 7 + 4

147+ 3 e7t

490e7t490 ndash endash 3t

30e t30

3 y = ndash endash 2t9e t9 + 4 et

94 et9 + 2 32 3 endash t 22 cos 3t 23t 2 ndash 1 31 3 sin 3t 23t 2

4 x = endash t 2t 2 cos 15t 215t 2 ndash 1 151 15 sin 15t 215t 2

5 y = ndash et cos h 2t + et sin h 2t 2t cos h 2t + et sin h 2t 2

6 y = 19

51e4t + 11

15endash 2t ndash 9

85cos t ndash 2

85sin t

7

y = ndash t

2cos t + 1

2sin t

8 R = 1

5endash t cos h 2t + 3

2sin h 2t + 3 sin t ndash cos t

9 y = 43

10endash t ndash 29

13endash 2t ndash 9

130cos 3t ndash 7

130sin 3t

10 y = ndash

endasht

102 cos t + sin t + 6

5cos h 2t ndash 1

20sin h 2t

11 y = 2 e2t ndash 3 endash t ndash endash t sin 3t + endash t cos 3t 12 y = endash2t ndash endash3t ndash 2tendash3t

13 y = ndash4tet+ 3e

tndash 3 cost + sint 14 y = ndash

2et

5cos t +

et

5sin t + sin t

5+ 2 cos t

5

15 y = t ndash τ

0

tendash 2 t ndash τ g τ dτ

όπου g η τριγωνική κυματική

συνάρτηση του Σχήματος 1)

16

y1 = et + e2t

y2 = e2t

17

y1 = et

y2 = endash t

y3 = et ndash endash t

18 x = 8 sin t + 2 cos t

y = ndash 13 sin t + cos t + (et ndash endasht)2 19

x = endash 2t

ndash t et

y = etndash e

ndash 2t+ 3t e

t 3

20 x = ndash e

t+ e

ndash t+ 4 e

2t 2

y = et

+ endash t

ndash 2e2t

2

21

x = 3 ndash 3 e2t 32t 3 + 2 t

2+ 2 t 4

y = ndash 3 e2t 32t 3 + 2t + 3 2

22 x = ndash 1 + 1 e

ndash 2t 2

y = 1 + 2 t ndash 1endash 2t

4

23 x = ndash endash t cos 2t ndash 32 endash t sin 2t

y = 7endash t cos 2t + 19 endash t sin 2t ndash 6

1

t

g(t)

1 2 3 4 5

Σχήμα 4 Τριγωνική κυματική συνάρτηση

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 42: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 45

24

x = ndash 21 ndash 18t + endash 2t + 20 et 6

y = ndash 3 + endash 2t + 2et 3 25

x t = ndash 79

ndash 13

t ndash 3199

endash 3t 53t 5 + 10899

et 2t 2

y t = ndash 139

ndash 13

t + 6299

endash 3t 53t 5 + 8199

et 2t 2

26 x t =

1 ndash ttt

e2t

27

x t =

t ndash t2 ndasht3

6ndash

t4

2+

t5

2ndash t22

t2 + t36

28

x t =

11 + t

11 + 2t

e3t

29 x t = 2 et t cos t + 3t sin t + sint

ndash t sin t

33 y x =

w0

24 E Ix2 x ndash L 2

34

y x =

w0

E IL2

4x2 ndash L

6x3 + 1

24x4

35 (i) x t = a1 endash Bt 2MBt 2M cos κ Mκ M ndash B

2

4 M2B

2

4 M2

t

+2 a2 M + B a1

4Mκ ndash B2

endash Bt 2MBt 2M sin κ Mκ M ndash B

2

4 M2B

2

4 M2

t

36 I1 = ndash endash 20t ndash 2 endash 5t + 3 I2 = ndash 2 endash 20t + 2 endash 5t I3 = endash 20t ndash 4 endash 5t + 3

76 Προβλήματα

1 f t = H1 t + H3 t 2 f t = 5 + 2 t ndash 3 H3 t

3 f t = endash t ndash endash t H2 t 4 f t = 4 H2 t ndash H5 t

5 f t = cos t ndash cos t Hπ 2π 2

t 6

f t = t H1 t ndash H 2 t

7 f t = 4 ndash 2 H1 t ndash 2 H2 t

8 f t = t ndash 1 H1 t + H1 t ndash t

9 f t = e6 H2 t e3 t ndash 2

11

F s = 48

endash 2s

s

12 F s = endash 4s 128

s + 96s2

+ 48s3

+ 12s4

13 F s = ndash

endash 5s

s2ndash

3endash 5s

s + 2s2

14 F s = h

s1 ndash endash 4s

1 + endash 4s

15 F s =

1000 endash5 s

s +600 endash5 s

s2+

240 endash5 s

s3+

48 endash 5s

s4ndash endash3 s + 4 3s + 13

s + 4 2

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError 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false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 43: 978-960-7182-07-4

46 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

16 F s = 1

s2ndash

37 endash s

s ndash14 endashs

s2ndash

14 endashs

s3

17 F s =6320 endash 9s

s +2862 endash 9s

s2+

966 endash 9s

s3+

216 endash 9s

s4+

24 endash 9s

s5

18 F t = 3

s + 2 2 + 9endash s

19 F s =

s2 + s + 2

s3endash 2s

20 F s = 2 endashss32 e s3

21 F s =e3 ndash 3s

s ndash 1

22 F s = 1

s2ndash

endash s

s2ndash

endash 3s

s2+

endash 4s

s2 23

F s = 1

s2ndash

2 endash 2s

s2+

endash 4s

s2

24 F s = 1

s2ndash

2 endash s

s2 1 + endash 2s

25 F s = 1

s2ndash

2 endash s

s2+

endash 3s

s2+

endash 4s

s

26 F s = 1

s + ndash 1n

Σn = 0

infine

ndash ns=

1 s1 s

1 + endash s sgt 0

27

F s =

1 ndash 1 + s endashs

s2 1 ndash endashs

28 F s =

1 + endashsπ

1 + s2 1 ndash endashsπ s gt 0

29 f t = 12

H1 t sin 2t ndash 1

30 f t =

endash t

5sin 5 t ndash Hπ t

endash t ndash π

5sin 5 t ndash π

31 f t = H05 t ndash endash5 t ndash 1 21 2 + e t ndash 1 21 2

6t 1 21 2 + e 1 21 2

6

32 f t =

H2 t

7+ 3

7ndash 1

7H2 t endash2 t ndash2 cos 3 t ndash 2

+ 23

sin 3 t ndash 2 ndash 37

endash 2 t cos 3 t + 23

sin 3t

33 f t = 5 e4t + 4t e4t ndash 5 e5t 34 f t = ndash 27 1 + t endash 3t + 28 ndash 8 t endash 2t

35 f t = ndash 1

16e2t + 1

2t e2t + 1

16endash 6t

36

f t = 8

25endash 2t + 42

25e3t + 18

5t e3t

37 f t = 41

7endash 3t ndash 13

7endash 3t 23t 2 cos 19 t

5ndash 129 19

133endash 3t 23t 2sin 19 t

12

38 f t = H2 t sin h 2 t ndash 2

39 f t = H1 t + H2 t ndash H3 t ndash H4 t

41

t = t et 2t 24t et 2t 24

42 f t = 1

4t ndash 1 e

t ndash 1 2t 1 2 H1 t

43 f t = 1

2et 2t 2 H2

t 2t 2 44

f t = 1

4t4

3endash 3 t 4t 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError 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HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 44: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 47

45 f t = 16

et 3t 3 e 2t 3t 3 ndash 1

46 f t = 2

0

tendash 5 p

2p

2 sin p2

p2 psin p2

p2 p dp

49 w = endash x ndash 2 2

50 F s = 3 eπs ndash 1 s2 + 1 eπs ndash 1ndash1

51 y t = 2 + 3t endasht + 2 H3 t t ndash 5 + t ndash1 endash tndash 3

52 y t = 3 cos t ndash sin t + 1

2t sin t + 1

2Hπ 2π 2

t t ndash π2

sin t ndash cos t

53 y t = t et + H1 t 2 + t + 2t ndash 5 et ndash 1 ndash H2 t 1 + t + 2t ndash7 et ndash 2

54 w = 1

5+

4 endash t

5cos 2t +

2 endasht

5sin 2t +

Hπ t

51 ndash endash tndash π cos 2 t ndash π +

sin 2 t ndash π2

55 R x = 1

2ndash

endash 2x

2ndash H1 x 1 ndash e

ndash 2 x ndash 1+ H2 x 1

21 ndash e

ndash 2 x ndash 2

56 y(t) = cos t + H1 t 1 + t ndash 1 ndash cos t ndash 1 ndash sin t ndash 1

57 N t = 1 ndash e

ndash tndash t e

ndash t+ t

3 endash t

3ndash H2 t 1 ndash e

ndash tndash 2ndash tndash 2 e

ndash t ndash 2

ndasht ndash 2 2

2endash tndash 2 ndash

t ndash 3 3

6endash tndash 2

58

y = endash t sin t + 12

Hπ t 1 + endash tndash π cost + endash tndash π sint

ndash 12

H2π t 1 ndash endash tndash 2π cost ndash e ndash tndash 2π sint

59 y = 1

61 ndash H2π t 2 sin t ndash sin 2t

60 y = 16

2 sin t ndash sin 2t ndash 16

Hπ t 2 sin t + sin 2t

61 y = 1 + ndash 1 n Hnπ t 1 ndash cos t ndash nπΣ

n = 1

infin

63 z1 = 1

18ndash 1

10cos 2t ndash 2

45cos 3t ndash

118

ndash 110

cos 2t ndash 2 ndash 245

cos 3t ndash 3 H1 t

z2 = 29

+ 445

cos 2t ndash 1445

cos 3t ndash

29

+ 445

cos 2t ndash 2 ndash 1445

cos 3t ndash 3 H1 t

64 I t = 2ω2

1 ndash cos ωt + 4ω2

ndash 1 nΣn = 1

infinHn t 1 ndash cos ω t ndash n

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true 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true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 45: 978-960-7182-07-4

48 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

77 Προβλήματα

5 y t = cos h t

2ndash 3 sin h t

2e3t 23t 2 ndash 2 H2 t sin h

t ndash 22

e3 t ndash 2 2

14 i t = endash R t Lt L

LL όχι

16

y t = 1

2sin t ndash t cos t ndash Hπ t sin t

17 y t = cos 3

2t + sin 3

2t + 8

3H1 t sin 3

2t ndash 1 ndash 4

3H2 t sin 3

2t ndash 2

18 y t = 3 t endash t + 12

t2 endash t + 3 H3 t t ndash 3 endash tndash 3

19 y = 2t endash t + H2π t 1 ndash endash tndash 2π ndash tndash 2π endash tndash 2π

20 y= 22

endash t sin 2 t +endash t

4cos 2 t+ 1

4sin t ndash cos t + 2

2Hπ t endash tndash π sin 2 t ndash π

21 y = cos ωt ndash ωndash1 Hπ ωπ ω t sin ωt

22

y = 1 + Hπ t sin t

23 y = 2 H1 t e2 t ndash 1 t ndash 1 ndash H2 t e2 t ndash 2 t ndash 2

24 N = π H2 x sin h x ndash 2 ndash π H4 x sin h x ndash 4

25 w = 2 cos t ndash H 3π

2t sin t ndash 3π

2 26 R = πHπ x sin3 x ndashπ = ndashπHπ x sin 3x

27 y t = P

6EI3a ndash x3 + x ndash a 3 Ha x

78 Προβλήματα

1

116

sin h 2t ndash sin 2t

2 cos at

sin a ndash b t

2 a ndash b+

sin a + b t

2 a + b+ sin at

sin a ndash b t

2 a ndash bndash

sin a + b t

2 a + b

3

1

a2 + b2cos h bt ndash cos at

4

cos h at ndash cos h bt

a2 ndash b 2

5

1a4

1 + t2

cos at ndash 32a5

sin at

6 ndash 1

5endash 2t + 1

3e3t + 1

6endash 3t

7 12

H4 t ndash 12

H4 t endash 2 t ndash 4

8

56

e5t ndash 56

e3t cos 8 t ndash 53 8

e3t sin 8 t

9 1

n ndash 1 tn ndash 1 eat

10 116

sin 2t ndash t8

cos 2t

11 23

endash 2t + 13

et

12

24 s4 s2 + 16 ndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError 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HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 46: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 49

13

4 ndash s2 s2 + 1 s2 + 4 2 ndash 1

14 Γ 3 23 2

s3 23 2

d3

ds3s

s2 + 1=

ndash 3 π s4 ndash 6s2 + 1

s3 23 2 s2 + 14

15 s

2

s2

+ a2

s2

+ b2

16

a2

s2

+ a2 2

17

f a0

tda

18

0

tf a da dr

0

r= t f a

0

tda ndash a f a

0

tda

19 Lndash1 F s

s2 + a2 20

Lndash1 F s

s2 ndash a2

24 Γ 4

3Γ 8

3t3

6 26

πt4

64

28 c I1 21 2

p t =aj j

Γ j + 3 23 2Σ

j = 0

ntj + 1 21 2

29 (c)Iv p t =Σ

j = 0

naj j Γ j + v + 1 tj + v

(d)

Iv eat = an Γ n + v + 1an Γ n + v + 1Σ

n = 0

infintn + v

30 (i) H (D12p)(t) δεν υπάρχει στο t = 0 αν a0 = 0

D1 21 2 p t = aj j Γ j + 1 21 2aj j Γ j + 1 21 2Σ

j = 0

ntj ndash 1 21 2

31 y = 142

f t lowast e3t ndash 142

f t lowast endash 3t ndash 228

f t lowast e 2 t ndash 228

f t lowast endash 2 t

32 y = 1

15f t lowast e3t ndash 1

15f t lowast cos 6t ndash 1

5 6f t lowast sin 6t

33 y= 1

10f t lowast e9t ndash 1

10f t lowast endash t ndash 1

5e9 t + 1

5endash t

34 y t = ndashe3t lowast f t

11+

e4t lowast f t18

ndash7 cos 2t lowast f t

198ndash 4 2

18f t lowast sin 2t

35 y = ndash 1

2f t lowast endash 6t + 1

2f t lowast endash 4t + endash 6t ndash e ndash 4t

36 y = endash tndash r

0

tsin t ndash r sin ar dr

37 y = 1

21 ndash endash t sin t ndash endasht cos t ndash 1

2H2π t 1 ndash endash t ndash 2π sin t ndash endash t ndash 2π cos t

38 y t = ndash 12

sin 2t + sin t

39 y t = 02 endash t + cos 5t ndash 2 sin 5t

40 y t = 1 ndash endash 2t 2 για 0 lt t lt 1 και

2y t = endash t + 1 ndash 1 1 + e2 endash 2t για t gt 1

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
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Page 47: 978-960-7182-07-4

50 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

79 Προβλήματα

Nα λυθούν οι ακόλουθες ολοκληρωτικές εξισώσεις με χρήση μετασχημα-

τισμού Laplace

1 y = 1

2δ t + 3

2sin t

2

y = 3 π t1 61 6 4 Γ 4

3Γ 7

6

3 y = 23δprime t (υποδ Lδ΄ = s) 4 y = πndash1 δprime t + π

5 y = tndash 1 21 2πt 1 21 2π endash 2t + 2 π2 π e r f 2t 6

y = δ t + Jprime0 t

7 y = 2t ndash 3

2sin t

8

y t = ndash 12

25endasht + 4

5t endasht + 12

25cos 2t + 9

25sin 2t

9 y (t) = t

10 y = t3 et

3et3 + t2 endash t + t endash t

11

y t = cos h t

12 y t = ndash

endash t

2+ 3

2cos t + 1

2sin t

13 y t = 2 sin h t

14

t = ndash 18

+ 94

t2 + 18

cos h 2 3 t

15

y t equiv 0

16 y t = 5

4et ndash 1

4endash t ndash 1

2t endash t

17

y t = endash t 2t 2 cos 15

2t ndash 1

15sin 15

2t

18 y t = ndash 12

et +endasht

2ndash t endash t

19 y t = endash t 20 y t = t + 32

sin 2t

21 y t = 2 ndash 2 cos t 22 y t = 3 23 y t = cos t + sin t ndash 1

24 y t = cos h t 25 y t = 38

e2t + 18

endash 2t + 12

cos 2t + 14

sin 2t

26 y t = endash t 2t 2 cos 3 t

2ndash 1

3endash t 2t 2 sin 3 t

2 27

y t = sin t ndash 1

2t sin t

28 i t = 20000 t endash 100t ndash tndash 1 endash 100 t ndash 1 H1 t

29 i t = endash 10t ndash endash 100 t + 1ndash e10 endash 10tndash 1ndash e100 endash 100 t H1 t

30 i t = 1 ndash H2π t sin 100 t

31 i t = 50ndash1 1 ndash endash 50 t 2

ndash 50ndash 1 H1 t 1 + 98 endash 50 t ndash 1 ndash 99 endash 100 t ndash 1

32 y t =

endash 3t

5+ 4

5e2t

33

y t = 1

2t2

34 y t = 12

sin t + 1 ndash 32

t endash t 35 Y s = s2 + 4 s2 s2 + 4 ndash 1ndash1

36

Y s = s s2 ndash 3s ndash Γprime 1 ndash ln sndash1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 48: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 8

82 Προβλήματα

6 (k π 0) όπου k = 0 plusmn1 plusmn2 9 (ndash2 0) (0 0) (1 0)

11 0 0 1 0 0 2 12 1

2 13 0 0 0 1 1 0 12 1

4

14 0 0 0 cd

ab

ndashc2

d2ndash

a2

b2 15 (0 ndash2 k π) k = 0 1 2

17 y = cos (x ndash 1)2 19 y = tan x

21 Τα σημεία x = x0 y = ndash1 ndash x0 και οι κύκλοι x2 + y2 = c2 εξαιρουμένων

των παραπάνω σημείων

23 y = c

x2ndash x

26 (x0 0) (0 y0) όπου x0 y0 αυθαίρετα και οι καμπύλες

y = c exp ndash 23

e3x x ne 0

Mε τη χρήση της ανάλυσης προσήμου να περιγραφεί η συμπεριφορά των

μη-στάσιμων λύσεων των ακολούθων εξισώσεων

27 ndash1 ασταθές 0 ασ ευσταθές 2 ασταθές 28 0 ασταθές

30 ndash2 0 2 ασταθή ndash1 1 ασ ευσταθή 32 0 ασ ευσταθές 2 ασταθές

34 2 ασ ευσταθές 3 ασταθές

83 Προβλήματα

9 (1 1) εστία ΑΕ 10 (ndash2ndash6) νόθος κόμβος Α

11 (ndash1 1) κόμβος Α 12 (31 ndash22) κέντρο Ε

13 όλα τα σημεία της ευθείας y=x2 (λ1=0 λ2=ndash3) ΑΕ

14 (2 ndash1) κόμβος Α 15 (0 0) σάγμα Α

17 (0 0) κέντρο Ε 19 (0 0) κέντρο Ε

21 (ndash1 1) σάγμα Α 22 (2 1) κόμβος ΑΕ

23 (2 2) σάγμα Α 24 (3 2) εστία ΑΕ

25 (ndash2 ndash1) κέντρο Ε

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 49: 978-960-7182-07-4

52 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

84 Προβλήματα

1 (0 0) εστία Α (1 1) σάγμα Α 2 (ndash1 0) σάγμα Α (1 0) κέντρο Ε

3 (ndash1 0) κόμβος ΑΕ (1 1) εστία Α (0 0) σάγμα Α

4 (ndash1 ndash1) εστία ΑΕ (4 4) εστία Α

5 (0 2) (0 ndash2) κέντρα Ε (2 0) (ndash2 0) σάγματα Α

6 (0 0) κέντρο Ε x = plusmn1 ευθείες κρίσιμων σημείων

7 (nπ mπ) κρίσιμα σημεία (i) n+m=άρτιος σάγμα A (ii) n+m=περιττός

κέντρο E

8 mπ 0 m = 2k νόθος κόμβος A

m = 2k + 1 σάγμα A

9 (0 0) μη-απλό κρίσιμο σημείο (2 ndash2) σάγμα Α 10 (0 0) εστία ΑΕ

11 (1 1) σάγμα Α (ndash1 ndash1) εστία ΑΕ 12 (2 1) σάγμα Α (ndash2 ndash1) εστία ΑΕ

13 (0 0) σάγμα Α (4 0) κόμβος Α (0 23) κόμβος ΑΕ (1 1) σάγμα Α

14 (0 0) κέντρο Ε (1 ndash1) σάγμα Α

15 (0 0) κέντρο Ε (1 1) σάγμα Α (ndash1 ndash1) σάγμα Α

16 (0 0) ασταθές για μgt0 και ασυμπτωματικά ευσταθές για μlt0

18 (0 0) κέντρο Ε (του γραμμικοποιημένου)

19 (0 0) κόμβος Α κόμβοι ΑΕ (plusmn1 0) κόμβοι ΑΕ σάγματα Α

20 Έστω Δ = a1b2 ndash b1a2 (i) Αν Δgt0 (0 0) σάγμα Α (a1a2 0) σάγμα Α (b1

b2 Δa3b2) ΑΕ κόμβος αν Δgtb1a224b2 και ΑΕ εστία αν Δltb1a2

24b2 (ii) Αν Δlt0

(0 0) σάγμα Α (a1a2 0) κόμβος ΑΕ iii) Αν Δlt0 τότε Υ(t)0 ενώ αν Δgt0

Y(t) Δa3b2

85 Προβλήματα

1 V = xy ασταθής 2 V = 15 x2 + 6xy + 3 y2 ασυμπτωτικά ευσταθής

3 V = xy ασταθής 4 V = 8 x2 + 11xy + 5y2 ασυμπτωτικά ευσταθής

5 V = 3 x2 + 14xy + 8y2 ασταθής 6 V = x2 + y2 ασυμπτωτικά ευσταθής

7 V = x2 + y2 ασυμπτωτικά ευσταθής 8 V = x2 + y2 ευσταθής

9 V = x2 + y2 ευσταθής 10 V = 3 x y2 ndash x3 ασταθής

11 V = x ndash y ασταθής 15 V = x2 + y2 ασυμπτωτικά ευσταθής

19 a = 2 m = 1 b = 1 n = 2 ασυμπτωτικά ευσταθή

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 50: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 53

20 a = b = 1 m = n = 2 ασυμπτωτικά ευσταθής

24 V = x2 + y2 S = IR2 25 V = x2 + y2 S = IR2

26 V = x2 + y2 S =(x y) x2 + y2 lt 1

27 V = (xa)2 + (yb)2

S = (x y) (xa)2 + (yb)2 lt 1

86 Προβλήματα

1 Δ4 = 0 επομένως ασταθής 2 Δ2 = 0 επομένως ασταθής

5 ευσταθής 7 όχι ευσταθής

8 ευσταθής 9 ευσταθής

11 Για ασυμπτωματική ευστάθεια θα πρέπει να έχουμε

0 0 0 0

1 0 0 1 0

1 0 b 1 0 b

0 b 0 c 0 b0 c 0

0 0 0 c

gt gt gt gt

ή

αgt0 αbgt0 ndashα2cgt0 ndasha2c2gt0 αδύνατον

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
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  • Κεφάλαιο 10
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true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars 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Page 51: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 9

91 Προβλήματα

1 f x ~ sinh612

+ Σn = 1

infin ndash 1n

e6 ndash e ndash 6

36 + n2π23cos nπx

3ndash nπ

6sin nπx

3

2 f x ~2π2

3+ Σ

n = 1

infin 4 ndash 1n

n2cos nx +

2 ndash 1n

n sin nx

3 f(x) ~ ndash 4 an = bn = 0 n = 1 2

4 f x ~ 12

+ Σn = 1

infinndash 2nπ sin nπ

2cos nπx

2+ 4

nπ cos nπ2

ndash cos nπ sin nπx2

5 f(x) ~ ndash 1 an = bn = 0 n = 1 2 6 f x ~4π2

3+Σn = 1

infin4n cos nx ndash 4π

n sin nx

7 f x ~

π2

6+ Σ

n = 1

infin 2 ndash 1n

n2cos nx + ndash 1

nndash 1 2

πn3ndash

ndash 1n π

n sin nx

8 f x ~ 2π ndash Σ

n = 1

infin4

π 4n2 ndash 1cos nx b n = 0 n = 1 2

9 f x ~ 12

+ 12

cos 4x an = 0 για n ne 0 4 και b n = 0 n = 1 2

10 f(x) ~2sin (3x) (δηλαδή η f(x) είναι μια σειρά Fourier στο [ndashπ π])

11 f x ~ ndash 13

sinh 3 + Σn = 1

infin 6 ndash 1n+1

sinh 3

9 + n2π2cos nπx

3

+ ndash 18nπ + 2nπ

9 + n2π2sinh 3 ndash 1

nsin nπx

3

12 f x ~ ndash 12

+ Σn = 1

infin6

n2π21 ndash cos nπ cos nπx

3equiv

equiv ndash 12

+ Σn = 1

infin12

2n ndash 12 π2

cos2n ndash 1 πx

3

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 52: 978-960-7182-07-4

56 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

92 Προβλήματα

1 λεία 2 τίποτα από αυτά

3 λεία 4 τμηματικά λεία συνεχής

5 τμηματικά συνεχής 6 τμηματικά λεία συνεχής

7 συνεχής όχι τμηματικά λεία 8 τίποτα από αυτά

9 ναι L = 2π3

10 ναι L = 1

11 ναι L = 6 12 ναι L = π

13 ναι L = 12 14 όχι

15 όχι 16 ναι L = π2

17 ναι L = π6

18 όχι

19 ναι L = π2

20 ναι L = 2π

21 άρτια 22 περιττή

23 άρτια 24 άρτια

25 περιττή 26 τίποτα

27 περιττή 28 άρτια

29 τίποτα 30 άρτια

31 περιττή 32 τίποτα

33 άρτια 34 άρτια

37 x x x xe e e e

g(x) = 2 2

38 g(x) = x2 (1 ndash x2)ndash1 + x(1 ndash x2)ndash1

39 g(x) = 3 + (x5 ndash 7x) 40 g x =2 1 + x2

1 ndash x2+ 4x

1 ndash x2

41 f(x) 0 στο IR 45 (a) 12 (b) 52 (c) 2 (d) 12 (e) 1

48 f x = x2 0 le x lt 3 f x = x ndash 4 2 3 lt x le 4 f 3 = 5

49 f x = sinx 0 lt x lt 2 f 0 = f 2 = sin22

50 f ndash 2ndash = ndash 4 f ndash 2+ = 0 f 1ndash = 0 f 1+ = 1 f ndash 3+ = ndash 6 f 3ndash = 9

fαprime ndash 2 = 0 fδ

prime ndash 2 = 0 fαprime 1 = 0 fδ

prime 1 = 2 fδprime ndash 3 = 2 fα

prime 3 = 6

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 53: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 57

f x = x2 1 lt x lt 3 f x = 0 ndash 2 lt x lt 1 f x = 2x ndash 3 lt x lt ndash 2

f 3 = f ndash 3 = 32

f 1 = 12

και f ndash2 = ndash 2

51 f 0ndash = 0 f 0+ = 0 fδ

prime 0 = fαprime 0 = 0 f ndash π+ = π2

f πndash = 2 fδprime ndash π = ndash 2π fα

prime π = 0 f x = x2 ndash π lt x lt 0 f x = 2 0 lt x lt π

f 0 = 1 f ndash π = f π =π2 + 2

2

52 f 0ndash = 1 f 0+ = 0 fδprime 0 = 1 fα

prime 0 = 0

f ndash π+ = ndash 1 f πndash = 0 fδprime ndash π = 0 fα

prime π = ndash 1

f x = cosx ndash πltxlt0 x = sinx 0ltxltπ f 0 = 12

f ndash π = f π = ndash 12

f

53 f 1ndash = 1 f 1+ = e fδprime 1 = e fα

prime 1 = 1

f ndash 2+ = ndash 2 f 2ndash = e2 fδprime ndash2 = 1 fα

prime 2 = e2

f x = x για ndash 2 lt x lt 1 f x = ex για 1 lt x lt 2

f 1 = 1 + e2

f 4 = f ndash 4 =e2 ndash 2

2

58 Οι συναρτήσεις 55 56 57

59 (i) f x ~ π2

+ 4π Σn = 1

infin cos 2n ndash 1 x 2n ndash 1 2

(ii) f x ~ 2 Σn = 1

infin sin nx n

(iii) f x ~ π2

+ Σn = 1

infin sin 2nxn

60 (a) f x ~ 1 + 8π2 Σn = 1

infin cos n ndash 12πx

2n ndash 1 2 (b) f x ~ 4

π Σn = 1

infin ndash 1nndash1

n sin nπx2

61 (a) f x ~ 2π + 4

π Σn = 1

infin cos 2nx

1 ndash 4n2 (b) f (x) = sin x

62 (a) f x ~e2 ndash 1

2+ 4 Σ

n = 1

infin e2 ndash 1n

ndash 1

n2π2 + 4cos nπx

(b) f x ~ 2π Σn = 1

infin n 1 ndash e 2 ndash 1n

n2π2 + 4sin nπx

63 (a) f x ~ 1π sinhπ + Σ

n = 1

infin2π

1

1 + n2ndash 1

nsinhπ cos nx

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 54: 978-960-7182-07-4

58 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

(b) f x ~ 2π Σn = 1

infinn

1 + n21 + ndash 1 n+1 coshπ sin nx

64 (a) f(x) ~3 (b) f x ~ Σn = 1

infin12

2n ndash 1 πsin 2n ndash 1 x

65 (a) f x ~π2 ndash 2

2π + Σn = 1

infinndash 4πn2

cos n ndash 2πn sin n + 2

πn21 ndash ndash 1 n cos nx

(b) f x ~ 2π Σn = 1

infinndash 2

n2sin n + 2

n cos n ndash πn ndash 1

nsin nx

66 (a) f x ~ 3sin 1 + 6sin 1 Σn = 1

infin1

1 ndash n2π2cos 2nx

(b) f x ~ Σn = 1

infin 12 2n ndash1 π cos 1

2n ndash12 π2 ndash 4

sin 2n ndash 1 πx

67 (a) f x ~ 12

+ 12

cos 2πx

(b) f x ~ 43π sin πx + Σ

n = 3

infin 1 ndash ndash 1 n

nπ +1 ndash ndash 1 n

n2 ndash 4 πsin nπx

68 (a) f x ~ 1 ndash Σk = 1

k ne 2m

infin4

k2π2cos kπx

(b) f x ~ 8 Σk = 1

infin sin 4k + 1 πx2

4k + 12 π2

+sin 4k + 3 πx

2

4k + 32 π2

69 (a) f x ~ 23

+ Σk = 1

infin3

3k + 2 πcos

6k + 4 πx3

ndash 33k + 1 π

cos6k + 2 πx

3

(b) f x ~ Σk = 1

infin2

6k + 1 πsin

6k + 1 πx3

ndash 46k + 3 π

sin 2k + 1 πx

93 Προβλήματα

1 π2

6= 1 6449341 2

π4

90= 1 0823

3 (β) πcos aπasin aπ

ndash 1a2

2 (γ) π2

a2sin2 aπ1 +

sin 2aπ2aπ ndash 2 andash 4 4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 55: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 59

94 Προβλήματα

1 (a) x2~π2

3+ 4 Σ

n = 1

infin ndash 1n

n2cos nx (b) x4 ~

π4

5 Σn = 1

infin48

ndash 1n+1

n4+ 8π2 ndash 1

n

n2cos nx

2 f prime x ~ Σn = 1

infin 2 ndash 1 n ndash 1

n2π2cos nπx

2+

2 ndash 1 n+1

nπsin nπx

2

3 y x = 1 + Σn = 1

infin Lannπ sin nπx

Lndash

Lb nnπ cos nπx

Lndash 1

95 Προβλήματα

1 f x y ~ Σn = 1

infin

Σm = 1

infin4π

2 ndash n2π2

n3ndash 1 n ndash 2

n3

ndash 1 m+1

m sin nx sin mx

2 f x y ~ Σn = 1

infin

Σm = 1

infin4mnπ2 e2 ndash 1 m+1 + 1

4 + m2π2

e4 ndash 1 n+1 + 1

16 + n2π2sin nπx

4sin

mπy2

3 f x y ~ Σn = 1

infin

Σm = 1

infin 16 ndash 1 n ndash 1 m

n 4 + m2π2m sinh 2 sin nπx

4sin

mπy2

4 f x y ~ Σn = 1

infin

Σm = 1

infinndash 8m

sin 8 ndash nπ8 ndash nπ ndash

sin 8 + nπ8 + nπ ndash 1 m sin nπx

4sin my

5 f x y ~ 2π2 + Σm = 1

infin8

m2ndash 1 m ndash 1 cos

my2

+ Σn = 1

infin8

n2ndash 1 n ndash 1 cos nx

2

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash1 n ndash 1 ndash 1 m ndash 1 cos nx

2cos

my2

6 f x y ~ π9

+ Σm = 1

infin4

3m2ndash 1 m cos mπy + Σ

n = 1

infin4

3n2ndash 1 n sin nx

+ Σn = 1

infin

Σm = 1

infin16

n2m2π2ndash 1 n+m cos nx cos mπy

7 f x y ~ 2 + Σ

n = 1

infin8

n2π2ndash 1 n ndash 1 cos nπx

4

8 f x y ~ 1009

+ Σn = 1

infin 400 ndash 1 n

3m2π2cos

mπy5

+ Σn = 1

infin

Σm = 1

infin1600

n2m2π2ndash 1 n ndash 1 m cos nπx

2cos

mπy5

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
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Page 56: 978-960-7182-07-4

60 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

9 f x y ~ Σn = 1

infin 4 sin h 1nπ ndash 1

n+1sin nπx

2

+Σn = 1

infin

Σm = 1

infin 8 ndash 1 n ndash 1 m

nπ 1 + m2π2sinh 1 ndash sin nπx

2cos mπy +mπ sin nπx

2sin mπy

10 f x y ~π4

9+ 1

2 Σm = 1

infin83π2 ndash 1

m

m2cos mx + 1

2 Σn = 1

infin83π2 ndash 1

n

n2cos ny

+ 12 Σn = 1

infin83π2 ndash 1

n

n2cos ny

11 f x y ~ Σm = 1

infin 2 ndash 1m+1

m sin mx sin y

96 Προβλήματα

1 f x ~ 100π 0

infinsin 2 ndash x α + sin αx dα

α

2 f x ~ 1π 0

infin cos αx + α sin αx1 + α2

3 f x ~ 2π 0

infin 1 ndash cosαα sin αx dα

4 f x ~ 4bπc2 0

infin 1α3

sin αc ndash αc cos αc cos αx dα

5 f x ~ 4π 0

infin sin α ndash α cos αα3

cos αx dα

6 f x ~0

infin cos xα + α sin xα1 + α2

dα 7 f x ~0

infin cos πα2

cos αx

1 ndash α2dα

8 f x ~ 2π 0

infin α3sin αx

α4 + 4dα

9 f x ~

0

infin sin πα sin xα1 ndash α2

10 f x ~ 2π 0

infinb2 ndash 2

b2sin αb + 2b

α cos αb cos αxα dα

11 f x ~ 2π 0

infin 1α cos

α b + c ndash 2x2

sinα c ndash b

2dα

12 f x ~ 2

π 0

infin 1α cos αb ndash cos αc sin αx dα

f x ~ ndash 2π 0

infin 1α sin αb ndash sin αc cos αx dα

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

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    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true 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Page 57: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 61

13 f x ~ 2bπc 0

infin 1 ndash cos αcα2

cos αx dα 14 f x ~0

infin 2απ 1 + α2

sin αx dα

15 I = 3π16

97 Προβλήματα

1 fS n = π ndash 1n 6

π3ndashπ2

n 2 fS n = n

n2 + 11 ndash ndash 1

neπ

3 fS n =

nn2 ndash a 2

1 ndash ndash 1 n cos aπ για anotinN

0 για aisinΝΝ

4 fS n = n

n2 + 11 ndash ndash 1

nendashπ 5 fC n =

2π ndash 1 n

n2 για n = 1 2 3

π3

3 για n = 0

6 fC n =ndash 1

ne ndash 1

1 + n2 7

Aν a = k isin ΝΝ fC n =

π2

για n = k

0 για n = 0

98 Προβλήματα

1 F f x = F α = b

α2 + b22π 2 F f x = F α = n

b ndash iα n+12

3 F f x = F α = 2α eiα b+c 2iα b+c 2 sin

α c ndash b2

4 F f x = F α = cα2b

4 sin2 bα22π

5 F f x = F α = ndash 4cα2b

cos αb + 4c

α3b2sin αb 1

6 F f x = F α = 2π

cos πα2

1 ndash α2eiπα 2iπα 2

7 F f x = F α = i ndash 1ndash cos aα

8 F f x = F α = 21 + α2

sinhL cos L(a) + 2α1 + α2

coshL sin L(a) 12π

9 F f x = F α = πendash α 12π 10 F α = 1

2cos

α2

4+ π

4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
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  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 58: 978-960-7182-07-4

62 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

11 F α = 1

2cos

α2

4ndash π

4 12 F α = 2

1 + α2ndash 1

1 + α ndash 2 2ndash 1

1 + α + 2 2 12π

13 F α = exp ndash 12α + 3i

2

214 F α = exp ndash 1

21 + α 2 + exp ndash 1

21 ndash α

212

15 FC α = 1α sin αc ndash sin αb FS α = 1

α cos αb ndash cos αc2π

16 FC α = 4cα2b

cos αd sin2 αb2

FS α = 4cα2b

sin αd sin2 αb2

17 FC α =1 ndash α2

1 + α2 2 FS α =

α1 + α2 2

18 FC α =2L cos αL

α2+

L2α2 ndash 2

α3sin αL 19 FC α =

b2 ndash α2

b2 + α2 2

20 FC α = π2 b

endash α24bα24b 21

FC α =

π2

endashα α gt 0

π2

eα α lt 0

22 FS α = 2α1 + α2 2

23 FS α = 12

1

1 + 1 ndash α 2ndash 1

1 + 1 + α 2

24 FS α =

0 για 0 lt α lt 1

π4

για α = 1

π2

για α gt 1

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
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  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 59: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

ΚΕΦΑΛΑΙΟ 10

102 Προβλήματα

1 y(x) 0 μοναδική 2 y(x) = c cos 3x

3 y x = c1 cos π x + c2 sin π x c1 c2 αυθαίρετα 4 y(x) 0 μοναδική

5 y(x) 2 sin (k x)sin k μοναδική αν k nπ n = 0 1 2 hellip

6 y x = 12 e

3 sin h x ndash cos h x μοναδική 7 y x = 3 x e3 x ndash 1 μοναδική

9 k = n π n = 1 2 3 y x = c sin n π x c αυθαίρετο

10 (i) bndasha = nπ n = 1 3 5hellip (ii) bndashanenπ n= 1 3 5 hellip (iii) bndasha = nπ n = 2 4 6

11 όχι λύση 12 y x = ndash 3 π23 π2 sin π x + 4x π4x π ndash2 μοναδική

13 y x = 1 51 5 endashx ndash 6 56 5 cos 2x μοναδική 14 y x = x ndash 1 endash x + x ndash 2

μοναδική

16 y(x) = c x3 sin (4 log x) c αυθαίρετο 18 y x = 1 ndash 12

log x x

20 y x = 4 54 5 x1 21 2 x ndash 1 23 y x = 2x ndash π sin 4x 44x 4

25 y (x) = x ex 27 y x = c1 +c2 ndash c 1

bx

28 y x = x π2 + 1 26π2 + 1 ndash x26 ymax = π2 + 1

3 23 29 3π2 + 1

3 23 29 3

29 yPrime = ndash sin x 0 lt x lt π y 0 = 0 y π = 0 καμία λύση

30 y(x) = sin x ymax = 1 31 y(x) = [1 ndash (2x ndash 1)4]192 ymax = 1192

103 Προβλήματα

1 λn = n2 π2 16 ϕn x = cos n πx 4 n = 0 1 2

2 λn =2n ndash 1 2 π2

4 ϕn x = cos

2n ndash 1 π x2

n = 1 2

3 λn =n2 π2

81 ϕn x = sin

nπ x ndash 19

n = 1 2

4 λn = n2 π2 + 14

ϕn x = ex 2x 2 sin n π x n = 1 2

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 60: 978-960-7182-07-4

64 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

5 λn = ndash n2 π2 ϕn x = endash x sin n π x n = 1 2

6 λn = 98

+n2π2

2 ϕn x = e3x 23x 2 sin n π x n = 1 2

8 λn = n2 π2 ϕn x = sin n π log x n = 1 2

9 λn =2n ndash 1 2π2

4 log 2 2 ϕn x = cos

2n ndash 1 π log x2 log 2

n = 1 2

10 δεν έχει πραγματικές ιδιοτιμές 12 λ0 = 0 φ0(x) = x ndash 1

13 λn = n4 φn(x) = sin (n x) n = 1 2

14 λn = k4n όπου cos h kn cos kn = 1 n = 1 2 hellip

ϕn x = sin h k n 1 ndash x + sin k n 1 ndash x + cos h k n sin k n x

ndash sin k n cos h k n x + cos k n sin h k n x ndash sin h k n cos k n x

15 (i) λ = ndash k2 lt 0 ϕ x = endash 2x sin h 3 k 2 ndash x sin h 6k3 k 2 x sin h 6k

(ii) λ = 0 ϕ x = 2 ndash x endash x2e 2 x2

(iii) λ = k2 gt 0 k ne n π 6π 6 ϕ x = endash 2x sin 3k 2 ndash x sin 6 k3k 2 x sin 6 k n isin NN

16 ω1 = 1 + 45 π24 ρ1 + 45 π24 ρ ϕ1 x = c xndash 1 21 2 sin π log x

18 (ii) ϕn x = 1 ndash cos 2n π x n = 1 2 3

ϕn x = sin k n ndash k n 1 ndash cos k n x ndash 1ndash cos k n sin k n x ndash k n

όπου tan (kn2) = kn2

104 Προβλήματα

1 x endash x yprime prime + k endash x y = 0 μ x = endashx

2 endash x2yprime prime + 2 k endash x2

y = 0 μ x = endash x2

3 1 ndash x2 yprime

prime+ k2 1 ndash x2 ndash 1 21 2 y = 0 μ x = 1 ndash x 2 ndash 1 21 2

4 yprime + λ xndash1 y = 0 μ x = xndash1 5 x endashx yprime prime + λ endash x y = 0 μ x = endash x

6 x2 + 1 yprime prime+λ xndash2 y = 0 μ x = xndash 2

7 sin x yprime prime+λ sin x y = 0 μ x = 1

8 endash x2 2 yprimeprime+ λ endash x2 2 y = 0 μ x = endash x2 2

9 endash x yprime prime + λ endash x y = 0 μ x = endashx 10 x2 yprime prime

+ λ x2 y = 0 μ x = x2

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 61: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 65

11 yPrime + λ xndash 2 y = 0 μ x = xndash 2 12 cn = 2 π2 π

13 c0 = 1 π1 π cn = 2 π2 π n ge 1

24 λn = 2n ndash 1 π 2 L 2 ϕn = cn sin 2n ndash1 π x 2 L n isin NN

25 λn = ndash tan L λn ϕn x = An sin x λn n = 1 2 3

26 λn = 1 + n2 π2 ϕn x = sin n π ln x xndash1 n = 1 2 3

27 λn = 14

+ n πln 3

2 ϕn x = (x + 2) sin n π ln 3 ln x + 2 n = 1 2

28 λn = 1 + 2nπ ln 2 2 12 ϕn x = sinn π ln 2 ln 1 + x

1 + x n ge 1

29 λ0 = 0 ϕ0 x = 1 λn = n2 + 94

ϕn x = e3x 23x 2 sin (nx) ndash 23

n cos nx n e NN

30 λn = 16 n2 ϕn x = sin 4n tanndash1 x n = 1 2 3

31 λn = n2 π2 ϕn x = cos n π x ψn x = sin n π x n = 1 2 3

32 λn = 4 n2 π2 ϕn x = cos 2 n π x ψn x = sin 2 n π x n = 1 2 3

33 λn = n2 ϕn x = sin n x ψn x = cos n x n = 0 1 2 3

34 λn = 4 n2 ϕn x = sin 2n x ψn = cos 2n x n = 0 1 2 3

35 λ gt 0 ϕλ x = sin λ ln x 36 λ gt 0 ϕλ x = cos λ x

37 λn = 2n 2n ndash 1 ϕn x = P2n x Pn πολυώνυμο Legendre τάξης n

38 λn = n2 ϕn x = Tn x Tn πολυώνυμο Chebysh ev τάξης n

39 λ0= 0 φ0(x) 1 λn η θετική ρίζα τάξης n της εξίσωσης Jprime0 λ = 0

(J0 συνάρτηση Bessel μηδενικής τάξης και πρώτου είδους)

44 a = π λ 3 k ndash 1 A

105 Προβλήματα

7 ϕ0 x = 1 π1 π ϕn x = 2 π2 π cos n x n = 1 2 3

8 λn = 1 + n2 π2L2π2L2

1 21 2 ϕn x = 2 L e n2 π2 + L22 L endashx

n2 π2 + L2

9 λn =2n ndash1 2 π2

4 log b 2 ϕn x = 2

log bcos

2n ndash 1 π log x2 log b

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 62: 978-960-7182-07-4

66 ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ

10 λn =

2n ndash 1 π2

2

ϕn x = 2 sin2n ndash 1 πx

2

11 λn = n2 π2 ϕn x = 2 ex sin n π x

12 f x = x ~ 2 ndash 1 n ndash 1 cos n x n2 π + ndash1Σn = 1

infin π― 2

13 f x ~ 2 n π 1 ndash b cos n πΣn = 1

infinsin n2 π2 + log2 b

ndash1nπ log x log b

14 f x ~ 8 n π 1 ndash ea 2a 2 cos n πΣn = 1

infinendash x 2x 2 sin n π x 4n2 π2 + a2a2 ndash1

a

15 f x ~ 1 + 2 ndash1 n + 1Σn = 1

infinsin n x nsin n x n

16 f x ~ 2 π e3x 23x 2 n 1 + ndash 1 n ndash 1 endash 3 23 2Σn = 1

infinsin n π x n2 π2 + 9 4n2 π2 + 9 4

ndash1

17 f x ~ 2 π2 π ndash 1 n ndash 1Σn = 1

infinsin n π log x nn π log x n

18 f x ~ 16 L2π316 L2π3 Σ

n = 1

infinπ ndash 1 n + 1 2n ndash 1 ndash 2 sin 2n ndash 1 πx 2L2n πx 2L 2n ndash 1

ndash3

106 Προβλήματα

1 y x = 2 ndash 1 n + 1 sin n π x n2 π2 ndash 2 n πΣ

n = 1

infin

2 y x = 2 2 cos λn ndash 1 cos λn x λn λn ndash 2 1 + sin2 λnΣn = 1

infin

(βλέπε Παράδειγμα 4 της Ενότητας 103)

3 y x = 8sin n π 2n π 2 sin n π x

n2 π2 ndash 2 n2 π2Σn = 1

infin 4 y x =

cos π x

2 π2+

2x ndash 1

2 π2+ c sin π x

5 y x = c sin π x ndashx sin π x

2 π 6 y x = 2π3

ndash 1 n ndash 1 sin n π x

n3Σn = 1

infin

7 y x = 1 + sin x +cos 1 ndash1 cos x

sin 1

8 y x = ndash1 nndash1 2 e + 2n ndash 1 πΣ

n = 1

infin2 cosh (xndash1)

cosh 1ndash

4 + 2n ndash 1 2 π2 sin

2n ndash 1 πx2

10 y x = 3 1 ndash x endashx ndash2 endash x

π1 ndash ndash 1 n e2 sin n π x

n n2 π2 + 4Σ

n = 1

infin

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice

Page 63: 978-960-7182-07-4

ΑΠΑΝΤΗΣΕΙΣ ΠΡΟΒΛΗΜΑΤΩΝ 67

11 y x = 12

x 3 ndash x 2 ndash8 x2

πsin 2 n ndash 1 π log x log 2

2n ndash 1 1 + 2n ndash 1 2 π2 log 2 2Σn = 1

infin

13

y x = ndashcos ln x

sin 2+ 8

π2

4 cos 2n ndash1 π log x 22n π log x 2

2n ndash 1 2 2n ndash 1 2 π2 ndash 4Σ

n = 1

infin e2

15 y x = c ndash cos π x π c isin IR2 16 y x = c sin 4 log x +sin 3 log x

7

17 όχι λύση

  • Κεφάλαιο 1
  • Κεφάλαιο 2
  • Κεφάλαιο 3
  • Κεφάλαιο 4
  • Κεφάλαιο 5
  • Κεφάλαιο 6
  • Κεφάλαιο 7
  • Κεφάλαιο 8
  • Κεφάλαιο 9
  • Κεφάλαιο 10
    • ltlt ASCII85EncodePages false AllowTransparency false AutoPositionEPSFiles true AutoRotatePages None Binding Left CalGrayProfile (Dot Gain 20) CalRGBProfile (sRGB IEC61966-21) CalCMYKProfile (US Web Coated 050SWOP051 v2) sRGBProfile (sRGB IEC61966-21) CannotEmbedFontPolicy Warning CompatibilityLevel 13 CompressObjects Tags CompressPages true ConvertImagesToIndexed true PassThroughJPEGImages true CreateJobTicket false DefaultRenderingIntent Default DetectBlends true DetectCurves 00000 ColorConversionStrategy CMYK DoThumbnails false EmbedAllFonts true EmbedOpenType false ParseICCProfilesInComments true EmbedJobOptions true DSCReportingLevel 0 EmitDSCWarnings false EndPage -1 ImageMemory 1048576 LockDistillerParams false MaxSubsetPct 100 Optimize true OPM 1 ParseDSCComments true ParseDSCCommentsForDocInfo true PreserveCopyPage true PreserveDICMYKValues true PreserveEPSInfo true PreserveFlatness false PreserveHalftoneInfo false PreserveOPIComments true PreserveOverprintSettings true StartPage 1 SubsetFonts true TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages false ColorImageMinResolution 300 ColorImageMinResolutionPolicy OK DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 300 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages false GrayImageMinResolution 300 GrayImageMinResolutionPolicy OK DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 300 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages false MonoImageMinResolution 1200 MonoImageMinResolutionPolicy OK DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 1200 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (US Web Coated 050SWOP051 v2) PDFXOutputConditionIdentifier (CGATS TR 001) PDFXOutputCondition () PDFXRegistryName (httpwwwcolororg) PDFXTrapped False CreateJDFFile false Description ltlt ENU ([Based on Press Quality(GRAM-4)] Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false BleedOffset [ 0 0 0 0 ] ConvertColors ConvertToCMYK DestinationProfileName (US Web Coated (SWOP) v2) DestinationProfileSelector UseName Downsample16BitImages true FlattenerPreset ltlt ClipComplexRegions true ConvertStrokesToOutlines false ConvertTextToOutlines false GradientResolution 300 LineArtTextResolution 1200 PresetName ([High Resolution]) PresetSelector HighResolution RasterVectorBalance 1 gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MarksOffset 6 MarksWeight 0250000 MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector UseName PageMarksFile RomanDefault PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ltlt AllowImageBreaks true AllowTableBreaks true ExpandPage false HonorBaseURL true HonorRolloverEffect false IgnoreHTMLPageBreaks false IncludeHeaderFooter false MarginOffset [ 0 0 0 0 ] MetadataAuthor () MetadataKeywords () MetadataSubject () MetadataTitle () MetricPageSize [ 0 0 ] MetricUnit inch MobileCompatible 0 Namespace [ (Adobe) (GoLive) (80) ] OpenZoomToHTMLFontSize false PageOrientation Portrait RemoveBackground false ShrinkContent true TreatColorsAs MainMonitorColors UseEmbeddedProfiles false UseHTMLTitleAsMetadata true gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [612000 792000]gtgt setpagedevice