HW # I 4. 1.3,4.1.11

,4.2.11

,4. 3.5

,43.8

,4. 4.3

4.1.30 Determine interval in which solutions are sure to existt ( t - Dy

" "te

t

y' 't 4 tay

-

- O

Solution :

Tete write this so the leading term has a coefficient of 1.

y' "

,y

"

9=0

The coefficients f¥,, , ¥f÷,are not defined at t -

- o,I

Thus,solutions exist on the interval taobao.DK

Verify the given functions are solutions of the differential

eg . & determine their Wronski an

y.

" 't y

'= O ; I

,cos #

,sin I t )

Solution :case I :

y =L

Then y'= O ⇒ y

' '= O ⇒ y

' ' '= O

SO y' "

t y'

= O to = O

case 2 :

y-

- cosetThen y

'= - sin HI ⇒ y

' '= - cos It) ⇒ y

' "= sin At

SO y' " t y

'= sin It ) - sin HI

Case 3 :

y-

- sin It )Then y

'= cos Lt) ⇒ y

' '= - Sin Lt) ⇒ y

' "= -

cos HI

SO y' "

t y'

= - cos HI t cos ft ) = O

The WRONSKI an is :

W =/ toI%nsF¥ stiff,I -

- o - f- cosHIGHcost th - offs in HD kilts in HI - o]= cos att ) t sin Htt-- I

So µ=I to ⇒ solutions are linearly independent

4 Find the general solution of

y' "

-

y"

-

y't

y-

- O

solutionThis is a homogeneous problem , so we'll solve using the characteristic

equation & finding its Roots

y' "

-

y"

-

y'

t

y-

- O ⇒ r3

- ra - r + 1=0⇒

- -

ra I r - H - I Cr - D= O factor by grouping )⇒ er - 1) ( ra - H = O⇒ ( r - 1) I r - 1) I rt 17=0

so

- " " '

* since we have a repeated Root,we must multiply the last term

by t

4.3.50 Determine the general solution of

y' "

- 4 y"

= It et

Solution :

Weise the method of undetermined coefficientsFirst

,We'll find The homogeneous solution

y' "

- 4 y'

to ⇒ r4

- 4y2=0 ⇒ r2 ( ra - 4) = o ⇒ raft 2) IF2) = O ⇒ no

,I 2

Thus

y= C

,t Ca t t Cz e

-

httCy

eatt Y

because Root O has multiplicity 2

Second,we'll And the particular solution Y .

We guess it has the form :

Y = At 2 t Bt t C t DetBut this has some duplicate terms with the horn .

solution . So we

multiply the polynomial terms by ta :

y -

- At 't BE + Ct2

t

DetNext

, we calculate derivatives

Y'

= 4A test 313ft 2C t t DetY

' '= 12 At 2+6 Bt t 2C t Det

Y" '

= 24A t t 613 +2 Det t DetY' "

= 24 At DetNOW we can solve LED = get :

y141

- 4 y' '

= t2 t et⇒ [24 At Det ] - 4G 2 Ata the Bt tact Det ] -

- ta t et⇒ f 48AT Edt E 2415ft tC24A - 8C ] t CD - 4DJ et = I Edt Ot to

TIEThus

- 48A = I ⇒ A = -448- 2413=0 ⇒ B -

- o

24A - 8C = O ⇒ -

Ya - 8C -

- O ⇒CI'46-

3D -

- I ⇒ D=-

43.

If"ge=neI¥gjI÷y=a+c±+ge*+a,e*t4as⇒oe

14.38J Determine the general solution of

y' 4)

t y' "

= sin At )

Solution :

Weil use the method of undetermined coefficientsFirst

,we find the homogeneous solution

.

y' "

t y' ' ' ⇒ r

4tr 3=0 ⇒ r3(rtD=o ⇒ r = O

,- I

so we have

y= C

,t Cat tCzt2 t Cy e

- tt Y

because O is a Root of multiplicity 3

Second,let's find Y

. We guess it takes the form :

Y = Asin ( 2 t ) t Bos ( 2h

Since neither of these terms are in the him .

Sol .

,it's the correct

foam . Now we can take derivatives.

Y'

= 2A cos Cat) - 2B sin fat)y " = - 4 Asin Cat - 413 cos Cat)

Y' "

= - 8A CosCat) t 813 sin Cat )

y" '

= 16 AS in At ) t 1613COSCat)

Plugging these in,we can solve LED = gas :

yI "

t y' "

Sin 12T)⇒ a 6A sin fat) t 1613 cos Cat)] tf 8A Cos Cat) t 813 sin fat )] = sin ( at )⇒ 46 At 813] sin fat ) t E 8 At 16 B) cos Cat) = Is in fat) to cos At )

Thus

16 At 813=1- 8 At 1613=0 → 1613=8 A ⇒ A = 2B

So the first eq becomes

16/2137+813=1 ⇒ 4013=1 ⇒ 13=440⇒ A = 2B = 21440) = "

20

Then Y --

singlet) t

0¥24

The general Solution is :y=c,tgttczt7cyettIosinGHt'oCosGt

4.4.30 Use the method of Variation of parameters to determine the

general solution of

y' "

-

2y"

-

y't 2 y

= e4T

SolutionFirst we find the horn . solution

y' "

- 2 y' '

-

y'

t2y-

- O ⇒ 2=o⇒ r Hr - 2) - I ( r - 2) = O I factor by grouping)⇒ ( r - 2) I ra - D= o

⇒ I r - 2) ( rt Mr - D=O⇒ r= - I

, 1,2So y E C

,e

- tt get t geht

Second,we know thesolution has the formy = U,e

- tt

U2 et xU , eat

whep.ieUiettuaetitujeteo

(1)

uifettusettujfdeat ) 121{uie-ttuje.tt us'l4e2tfe4t

(3)We must solve fosui.ua

'

.us':

(3) - (1) ⇒

3uje2t=e4t

⇒ uj

=3eat

So we can rewrite H )&( 2) as :

uie-ttujet-tge.tt

( I ){- uiettuylet =

- Fett ( I )

Then I IHI ) ⇒2ujet=- e4t ⇒ us's -

ya estII ) - (5) ⇒ 2u

,'e-t=tze4t⇒ u,'= '

he est

Integrating ,we get

:

U ,=Etc, U2=

- It tf Uz=eItG30 6 6

SO the general solution is

y=

uie-ttuzettuze2t-fgItcye.tt/-eaItCa)ettfefttcs)e2t4t

= ae-tt.ge#cze2ttfIzo-t6ttf)e=Ge-ttfaet.iCze2tte4tIzo

✓.