unit1 maths

7/31/2019 UNIT1 maths

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UNIT-I

ORDINARY DIFFERENTIAL EQUATIONS

Higher order differential equations with constant coefficients Method of variation ofparameters Cauchys and Legendres linear equations Simultaneous first order

linear equations with constant coefficients.

The study of a differential equation in applied mathematics consists of three phases.

(i) Formation of differential equation from the given physical situation, calledmodeling.

(ii) Solutions of this differential equation, evaluating the arbitrary constantsfrom the given conditions, and

(iii) Physical interpretation of the solution.

HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH

CONSTANT COEFFICIENTS.

General form of a linear differential equation of the nth order with constant

coefficients is

XyKdx

ydK

dx

ydK

dx

yd

nn

n

n

n

n

n

=++++

..............2

2

21

1

1

.. (1)

Where KnK

K ...............,.........21 , are constants.

The symbol D stands for the operation of differential

(i.e.,) Dy = ,dx

dysimilarly D ...,,

3

33

2

22 etc

dx

ydyD

dx

dy

y

==

The equation (1) above can be written in the symbolic form

(D XyKDK nnn

=+++ )..........11 i.e., f(D)y = X

Where f (D) = D nnn

KDK+++

...........

1

1

Note

1. = XdxXD1

2. dxXeeXaD

axax

=

1

3. dxXeeXaD

axax

=+

1

(i) The general form of the differential equation of second order is


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2

(1)

Where P and Q are constants and R is a function of x or constant.

(ii)Differential operators:

The symbol D stands for the operation of differential

(i.e.,) Dy = ,dx

dyD

2

22

dx

ydy =

D

1Stands for the operation of integration

2

1

DStands for the operation of integration twice.

(1)can be written in the operator formRQyPDyyD =++

2 (Or) ( RyQPDD =++ )2

(iv) Complete solution = Complementary function + Particular Integral

PROBLEMS

1. Solve (D 0)652 =+ yD Solution: Given (D 0)65

2=+ yD

The auxiliary equation is m `0652 =+ m

i.e., m = 2,3xx BeAeFC 32. +=

The general solution is given byxx BeAey 32 +=

2. Solve 0362

2

=+ ydx

dy

dx

yd

Solution: Given ( ) 0362 =+ yDD The auxiliary equation is 01362 =+ mm

i.e.,2

52366 =m

= i23

Hence the solution is )2sin2cos(3 xBxAey x +=

3. Solve (D 0)12 =+ given y(0) =0, y(0) = 1

RQydx

dyP

dx

yd=++

2

2


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Solution: Given (D 0)12 =+

A.E is 012 =+m

M = i

Y = A cosx + B sinx

Y(x) = A cosx + B sinx

Y(0) = A =0

Y(0) =B =1

A = 0, B = 1

i.e., y = (0) cosx + sinx

y = sinx

3. Solve xeyDD 22 )134( =+ Solution: Given xeyDD 22 )134( =+

The auxiliary equation is 01342 =+ mm

im 322

3642

52164 ===

( )xBxAeFC x 3sin3cos. 2 +=

P.I. = xeDD

2

2 134

1

+

= xe2

1384

1

+

= xe2

9

1

y = C.F +P.I.

y = ( )xBxAe x 3sin3cos2 + + xe29

1

5. Find the Particular integral of y- 3y + 2y = exx e2

Solution: Given y- 3y + 2y = e xx e2

xx eeyDD 22 )23( =+

P.Ix

eDD 23

121

+=

= xe431

1

+

= xe0

1

=x xeD 32

1

= x xe32

1

=x

xe


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4

P.I ( )xeDD

2

22 23

1

+=

= - xe2

264

1

+

= - x xeD2

32

1

= - x xe2

34

1

= - xe x2

P.I. = P.I 21 .IP+

= xxe + (- xe x2 )

= -x( xe + ex2 )

6. Solve xydx

dy

dx

ydcosh254

2

2

=+

Solution: Given xydx

dy

dx

ydcosh254

2

2

=+

The A.E is m 0542 =+ m

m = i=

22

20164

C.F = e ( )xBxAx

sincos

2+

P.I = ( )

+

++=

++

254

12cosh2

54

122

xx ee

DDx

DD

= xx eDD

eDD

++

+

++

54

1

54

122

=541541 +

++

xx ee

=210

xx ee

IPFCy .. +=

= e ( )xBxAx sincos2 + -210

xx ee

Problems based on P.I = ( ) axDf

oraxDf

cos)(

1sin

)(

1

Replace D 22 aby

7. Solve xy

dx

dy

dx

yd3sin23

2

2

=++


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Solution: Given xydx

dy

dx

yd3sin23

2

2

=++

The A.E is m 0232 =++ m

(m+1)(m+2) = 0

M = -1, m = -2

C.F = Ae xx Be 2

+

P.I = xDD

3sin23

12

++

= xD

3sin233

12

++(Replace D 22 aby )

= xD

D

Dx

D3sin

)73(

)73(

73

13sin

73

1

+

+

=

= xD

D3sin

)7()3(

7322

+

= xD

D3sin

499

732

+

= xD

3sin49)3(9

732

+

= xD

3sin130

73

+

= ( )xxD 3sin7)3(sin3130

1+

= ( )xx 3sin73cos9130

1+

IPFCy .. +=

Y = Ae xx Be 2

+ ( )xx 3sin73cos9130

1+

8. Find the P.I of (D xsin)12 =+

Solution: Given (D xsin)12

=+

P.I. = xD

sin1

12+

= xsin11

1

+

= x xD

sin2

1

= xD

xsin

1

2

= xdxx

sin2

P.I =-

2

cosxx


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9. Find the particular integral of (D xxy sin2sin)12 =+

Solution: Given (D xxy sin2sin)12

=+

=- ( )xx cos3cos2

1

= - xx cos2

13cos

2

1+

P.I1 =

+x

D3cos

2

1

1

12

= x3cos19

1

2

1

+

= x3cos16

1

P.I 2 =

+

xD

cos2

1

1

12

= xcos11

1

2

1

+

= xD

x cos2

1

2

1

= xdxx

cos4

= xx

sin

4

P.I = x3cos16

1+ xx

sin4

Problems based on R.H.S = e axeoraxaxax cos)(cos ++

10. Solve (D xeyDx 2cos)44 22 +=+

Solution: Given (D xeyD x 2cos)44 22 +=+

The Auxiliary equation is m 0442 =+ m

(m 2 ) 2 = 0

m = 2 ,2C.F = (Ax +B)e x2

P.I 1 =xe

DD

2

2 44

1

+

= xe2

484

1

+

= xe2

0

1

= x xeD

2

42

1

= x xe2

01


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= x xe22

2

1

P.I2

=x

DD2cos

44

12

+

= xD

2cos442

12

+

= xD

2cos4

1

=

x

D2cos

1

4

1

=8

2sin

2

2sin

4

1 xx=

IPFCy .. +=

y = (Ax +B)e x2

8

2sin x

Problems based on R.H.S = x

Note: The following are important

.......1)1( 321 ++=+ xxxx .................1)1( 321 ++++= xxxx ..............4321)1( 322 ++=+ xxxx ..............4321)1( 322 ++++= xxxx

11. Find the Particular Integral of (D xy =+ )12

Solution: Given (D xy =+ )12

A.E is (m 0)12 =

m = 1

C.F = Ae xx Be+

P.I = xD 1

12

= xD21

1

= [ ] xD 121 = ( ) xDD .........1 22 +++ = [ ]...........000 +++ x = - x

12. Solve: ( ) 3234 2 xyDDD =+


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Solution: Given 3234 2 xyDDD =+

The A.E is m 02 234 =+ mm

m 0)12( 22 =+ mm

0)1( 22 =mm

m = 0,0 , m = 1,1C.F = (A + Bx)e xx eDxC )(0 ++

P.I = 3234 2

1x

DDD +

=( )[ ]

3

22 21

1x

DDD +

= ( )[ ] 3122

211

xDDD

+

= ( ) ( ) ( )[ ]...............222113

22

222 ++ DDDDDD

D

= [ ] 34322

43211

xDDDDD

++++

= [ ]241861 232

+++ xxxD

=

+++ x

xxx

D24

2

18

3

6

4

1 334

=

2

24

6

18

12

6

20

2345xxxx

+++

= 2345

123220

xxxx

+++

IPFCy .. +=

y = (A + Bx)e xx eDxC )(0 ++ + 2345

123220

xxxx

+++

Problems based on R.H.S = xeax

P.I = xaDf

exeDf

axax

)(1

)(1

+=

13. Obtain the particular integral of ( xeyDDx 2cos)522 =+

Solution: Given ( xeyDD x 2cos)522 =+

P.I = xeDD

x 2cos52

12

+

=

( ) ( )( )1Re2cos

5121

1

2

=

+++

placeDbyDxDD

ex


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= xDDD

ex 2cos

52212

12

+++

= xD

ex 2cos4

12

+

= xex 2cos44

1+

= xD

xex 2cos2

1

= xdxxex

2cos2

P.I =4

2sin xxex

14. Solve ( xeyD

x

sin)2

22 =+

Solution: Given ( xeyD x sin)2 22 =+

A.E is (m 0)12

=+

m = -2, -2

C.F. = (Ax + B)ex2

P.I =( )

xeD

x sin2

1 22

+

= e( )

xD

x sin22

12

2

+

= e xD

x sin12

2

= e xx sin

1

12

= -e xx sin2

IPFCy .. +=

y = (Ax + B)e x2 - e xx sin2

Problems based on f(x) = x axxorax nn cos)(sin

To find P.I when f(x) = x axxorax nn cos)(sin

P.I = axxoraxxDf

nn cos)(sin)(

1

VDfdD

dV

DfxxV

Df

+=

)(

1

)(

1)(

)(

1

i.e.,( )

VDfDf

DfV

DfxxV

Df

=

)(

1

)()(

1)(

)(

1 '

[ ]V

DfDfV

DfxxV

Df 2

'

)()(

)(1

)(1 =


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15. Solve (xx xexeyDD 32 sin)34 +=++

Solution: The auxiliary equation is m 0342 =++ m

m= -1,-3

C.F =A xx Bee 3

+

P.I ( )xeDD

x sin)34(

121

++=

=( ) ( )[ ]

xDD

sin3141

12

++

= e( )

xDD

x sin2

12+

= e( )

xD

Dx sin41

212

+

+

= [ ]xxe x

sincos25

+

P.I( )

xDD

e x

3)3(4)3

12

3

2++++

=

= e( )

xDD

x

2410

12

3

++

= xDDe x

123

24

101

24

++

= xDex

12

5124

3

=

12

5

24

3

xe x

General solution is IPFCy .. +=

y = Axx Bee 3 + [ ]xx

e xsincos2

5+

+

12

5

24

3

xe x

16 Solve xxydx

dy

dx

yd cos22

2

=++

Solution: A.E : m 0122 =++ m

m = -1,-1

C.F =xeBxA + )(

P.I = )cos()1(

12

xxD +

=( ) ( )

( )xDD

Dx cos

1

1

1

)1(222

+

+

+


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11

=( ) ( )

( )xDDD

x cos12

1

1

22

++

+

=( )

( )xDD

x cos121

1

1

2

++

+

=2

sin

1

2 x

Dx

+

=2

sin xx

=1

sin

2

sin

+D

xxx

=( )

1

sin1

2

sin2

D

xDxx

=

2

sincos

2

sin xxxx +

The solution is y =xeBxA + )( +

2

sincos

2

sin xxxx +

17. Solve ( ) xyD 22 sin1 =+ Solution: A.E : m 012 =+

m = i

C.F =A cosx +B sinx

P.I = x

D

2

2sin

1

1

+

=

+ 2

2cos1

1

12

x

D

=

+

+x

De

D

x 2cos1

1

1

1

2

12

0

2

=

+ x2cos

3

11

2

1

= x2cos6

1

2

1+

=y A cosx +B sinx + x2cos61

21 +

18. Solve ( ) xexxxydx

yd++= 1sin

2

2

Solution: A.E : m 012 =

m = 1

C.F = A xx Bee +

P.I ( ) )(

)(

1

)(

)('

)(

11 V

DfDf

DfxxV

Df

==


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12

=( )

( )xDD

Dx sin

1

1

1

222

=2

sin

1

22

x

D

Dx

=( )

+

12

cos2

2

sin2D

xxx

=2

cos

2

sin xxx

P.I ( ) xexD

2

221

1

1+

=

=( )[ ]

)1(11

1 22

xD

ex ++

= e

( ))1(

2

1 22

xDD

x+

+

=( )

12

932 23 xxxex +

y = A xx Bee + +

( )12

932 23 xxxex +

19. Solve xxeydx

yd x sin2

2

=

Solution: A.E : m 012 =

m = 1 C.F = A xx Bee +

P.I =( )

xxeD

x sin1

12

= e( )[ ]

( )xxD

x sin11

12+

= e( )

( )xxDD

x sin2

12+

= e ( ) ( )

+

+ xD

D

xDDx

x

sin12

22

sin2

122

sin ,sin xx = 1,122

=== putD

= e( )

( )

+

x

D

Dx

Dxx sin

12

22sin

12

12

= e( )

( )( )

( )

+

+

+

144

sin22sin

41

2122 DD

xDx

D

Dxx

Put D 12 =

= e( ) ( )( )

++

+ x

D

DDx

Dxx sin

169

4322sin

5

21

2


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13

= e [ ]

+++

x

D

DDxx

xx sin169

628cos2sin

5 2

2

= e ( )( )

++

x

Dxx

xx sin25

214cos2sin

5

P.I = e ( ) ( )

++

25

cos2sin14cos2sin

5

xxxx

xx

Complete Solution is

y = A xx Bee + +

e ( )( )

++

25

cos2sin14cos2sin

5

xxxx

xx

METHOD OF VARIATION OF PARAMETERS

This method is very useful in finding the general solution of the second orderequation.

Xyadx

dya

dx

yd=++ 212

2

[Where X is a function of x] .(1)

The complementary function of (1)

C.F = c 2211 fcf +

Where c 21 ,c are constants and f 21andf are functions of x

Then P.I = Pf 21 Qf+

P =

dxffff

Xf

2

'

1

1

21

2

Q =

dxffff

Xf

2

'

1

'

21

1

IPfcfcy .2211 ++=

1. Solve xyD 2sec42 =+

Solution: The A.E is m 042 =+

m = i2

C.F = C xCx 2sin2cos 21 +

f x2cos1= f x2sin2=

f x2sin2'

1 = f x2cos2'

2=

f xxfff 2sin22cos2 222'

1

'

21 +=

= 2 xx 2sin2cos22

+

= 2 [ ]1 = 2

P =

dxffff

Xf

2

'

1

1

21

2


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14

= dxxx

2

2sec2sin

= dxx

x2cos

12sin

2

1

=

dxx

x

2cos

2sin2

4

1

= )2log(cos4

1x

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxx

2

2sec2cos

= dxx

x2cos

12cos

2

1

= dx21

= x2

1

P.I = Pf 21 Qf+

= )2log(cos4

1x (cos2x) + x

2

1sin2x

2. Solve by the method of variation of parameters xxy

dx

ydsin

2

2

=+

Solution: The A.E is m 012 =+

m = i

C.F = C xCx sincos 21 +

Here xf cos1 = xf sin2 =

xf sin'1 = xf cos'

2 =

1sincos 222'

1

'

21 =+= xxffff

P =

dx

ffff

Xf

2

'

1

1

21

2

= dxxxx

1)sin(sin

= xdxx2sin

=( )

dxx

x2

2cos1

= ( ) dxxxx 2cos21

= + xdxxxdx 2cos2

1

2

1


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15

=

+

4

2cos)1(

2

sin

2

1

22

1 2 xxx

x

= xxxx

2cos8

12sin

44

2

++

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxxx

1

)(sin)(cos

= xdxxx cossin

= dxx

x2

2sin

= xdxx 2sin21

=

4

2sin)1(

2

2cos

2

1 xxx

= xxx

2sin8

12cos

4+

P.I = Pf 21 Qf+

= xxxx

xxxxx

sin2sin8

12cos

4cos2cos

8

12sin

44

2

+

+

++

4. Solve (D xeyD 22 )44 =+ by the method of variation of parameters.Solution: The A.E is 0442 =+ mm

( ) 02 2 =m m = 2,2

C.F = ( ) xeBAx 2+

= xx BeAxe 22 + xxef

2

1 = xef

2

2 = xx exef 22'1 2 += ,

xef 2'2 2=

( ) ( )xxxx

exeeexffff22222'

12

'

21 22 +=

= 2x ( ) ( ) ( )222222 2 xxx eexe

= ( ) [ ]12222 xxe x

= ( )22xe = xe4

P=

dxffff

Xf

2

'

1

1

21

2

=

dx

e

ee

x

xx

4

22


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16

= xdx =

Q =

dxffff

Xf

2

'

1

'

21

1

= dxeexex

xx

4

22

= xdx

=2

2x

P.I = xxx ex

ex

ex 22

22

22

22=

y = C.F + P.I

= (Ax +B) ex2

+x

ex 2

2

2

5. Use the method of variation of parameters to solve ( ) xyD sec12 =+ Solution: Given xyD sec1

2=+

The A.E is 012 =+m

m = i

C.F = xcxc sincos 21 +

= 2211 fcfc +

xfxf sin,cos 21 ==

xfxf cos,sin '2

'

1=

1sincos22

2

'

1

'

21 =+= xxffff

P=

dxffff

Xf

2

'

1

1

21

2

= dxxx

1secsin

= dxxx

cos

sin

=

xdxtan

= log (cos x)

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxx

1seccos

= dx = x

P.I = 21 QfPf +

= xxxx sincos)log(cos + y = xcxc sincos 21 + + xxxx sincos)log(cos +


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17

6. Solve axyaD tan22 =+ by the method of variation of parameters.Solution: Given ( ) axyaD tan22 =+

The A.E is 022 =+ am

aim = C.F = axcaxc sincos 21 +

axfaxf sin,cos 21 ==

axafxaf cos,sin '2'

1 ==

)sin(sincoscos'

12

'

21 axaaxaxaxaffff =

= axaaxa 22 sincos +

= )sin(cos22 axaxa +

= a

P.I = 21 QfPf +

P=

dxffff

Xf

2

'

1

1

21

2

= dxa

axax

tansin

= dxaxax

a cos

sin12

=

dxax

ax

a cos

cos112

= ( )

dxaxaxa

cossec1

= ( )

+

a

axaxax

aa

sintanseclog

11

= ( )[ ]axaxaxa

sintanseclog12

+

= ( )[ ]axaxaxa

tanseclogsin1

2+

Q = dxffffXf

2'

1'

21

1

= dxaaxax tancos

= axdxa sin1

= axa

cos1

2

21. QfPfIP +=

= ( )[ ] [ ]axaxa

axaxaxaxa

cossin1

tanseclogsincos1

22

+


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18

= ( )[ ]axaxaxa

tanseclogcos1

2+

IPFCy .. +=

= axcaxc sincos 21 + ( )[ ]axaxax

a

tanseclogcos1

2+

7. Solve xydx

ydtan

2

2

=+ by the method of variation of parameters.

Solution: The A.E is 012

=+m im =

C.F = xcxc sincos 21 +

Here xfxf sin,cos 21 ==

xfxf cos,sin '2'

1 ==

1sincos 22'12'

21 =+= xxffff

P=

dxffff

Xf

2

'

1

1

21

2

= dxxx

1

tansin

= dxxx

cos

sin 2

= dxx

x

cos

cos1 2

= ( )dxxx

cossec

= xxx sin)tanlog(sec ++

Q =

dxffff

Xf

2

'

1

'

21

1

= dxxx

1tancos

= dxsin= xcos

21. QfPfIP +=

= )tanlog(seccos xxx + IPFCy .. +=

= xcxc sincos 21 + )tanlog(seccos xxx +

8. Solve by method of variation of parameters 144 22

'''+=+ xy

xy

xy

Solution: Given 144 2

2

'''+=+ xy

xyy

i.e., 24'''2 44 xxyxyyx +=+

i.e.,2422

44 xxyxDDx +=+ .(1)Put x = ze


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19

Logx = logze

= z

So that XD = D '

1''2= DDDx

(1) ( )[ ] ( ) ( )24'''

441zz

eeyDDD +=+ z

eeyDD z22 4'' 45 +=+

A.E is 0452 =+ mm

( )( ) 014 = mm 4,1=m

zz ececFC 24

1. +=

Here zz efef == 24

1 ,zz efef == '2

4'

1 ,zzz eeeffff 555'12

'

21 34 ==

21. QfPfIP +=

P=

dxffff

Xf

2

'

1

1

21

2

=[ ]

+ dx

e

eeez

zzz

5

24

3

= +

dxe

eez

zz

5

35

3

= [ ]dze z

+2

1

3

1

=

+

23

1 2zez

=zez 2

6

1

3

1

Q =

dzffff

Zf

2

'

1

'

21

1

=( )

+dz

e

eeez

zzz

5

344

3

= + dzeee

z

zz

5

68

31

= ( ) + dzeezz3

3

1

=

+

zz

ee

33

1 3

= zz ee3

1

9

1 3

zzzzz eeeeezIP

+

=

3

1

9

1

6

1

3

1. 342


20/26

20

= zzzz eeeze 2424

3

1

9

1

6

1

3

1

IPFCy .. +=

=zz

ecec 24

1 + +zzzz

eeeze2424

3

1

9

1

6

1

3

1

= ( ) ( ) ( )3

1

9

1

6

1

3

1)(

424

2

4

1 ++zzzzz eeezecec ( )2ze

= ( ) ( )29

log3

244

2

4

1

xxx

xecec zz ++

DIFFERENTIAL EQUATIONS FOR THE VARIABLE COEFFICIENTS

(CAUCHYS HOMOGENEOUS LINEAR EQUATION)

Consider homogeneous linear differential equation as:

)1(............................1

11

1 Xyadx

ydxa

dx

yda nx

nn

n

nn

=+++

(Here as are constants and X be a function of X) is called Cauchys

homogeneous linear equation.

)(.............. 011

1

1 xfyadx

dyxa

dx

yda

dx

ydxa

n

n

nn

nn

n =++++

is the homogeneous linear equation with variable coefficients. It is also known

as Eulers equation.

Equation (1) can be transformed into a linear equation of constant Coefficients

by the transformation.

xzorex z log)(, ==

Then

dz

dy

xdx

dz

dz

dy

dx

dyDy

1===

==dz

d

dx

dD ,

===dzdxD

dxdx

Similarly

=

=

dz

dy

xdx

d

dx

dy

dx

d

dx

yd 12

2

=dx

dz

dz

dy

dz

d

xdz

dy

xdz

dy

dx

d

xdz

dy

x

+=

+

111122

2

2

222

2 11

dz

yd

dz

dy

xdx

yd+=

dzdy

dzyd

dxydx =

2

2

2

2

2


21/26

21

( )1222 == Dx Similarly,

( )( )2133 = Dx

( )( )( )32144 = Dx and soon.

=xD

( )122 = Dx

( )( )2133 = Dx

and so on.

1. Solve )sin(log42

22 xy

dx

dyx

dx

ydx =++

Solution: Consider the transformation

xzorex z log)(, ==

=xD

( )122 = Dx

( ) )sin(log4122 xyxDDx =++ ( ) ( )zy sin412 =+

R.H.S = 0 : ( ) 012 =+ y

A.E : imm ==+ ,012

C.F = A cosz + B sinz

C.F = A cos (log x) +B sin (log x)

P.I = ( )zsin41

12+

= zzzz

cos22

cos4 =

P.I = )cos(loglog2 xx

Complete solution is:y = A cos (log x) +B sin (log x) )cos(loglog2 xx

y = ( ) )cos(loglog2)cos(loglog2 xxxxA

2. Solve xxydx

dyx

dx

ydx log24

2

22

=++

Solution: Given ( ) xxyxDDx log2422 =++ (1)Consider: xzorex z log)(, ==

=xD

( )122

= Dx (1): ( )( ) zey z=++ 241


22/26

22

zzey =++ 232

A.E : 0232 =++ mm

M = - 2, -1

C.F =12 loglog2

+=+ xxzz BeAeBeAe

C.F =xBA +

2

P.I =( )

( )zez23

12

++

=e( ) ( )

zz

2131

12

++++

=( )

zez

65

12

++

= z

ez1

2

6

5

16

++

=

=

6

5

66

51

6z

ez

e zz

=

6

5log

6

log

xe x

=

6

5log

6x

x

Complete solution is y = C.F +P.I

=x

B

x

A+2 +

6

5log

6x

x

3. Solve ( giventhatxyxDDx ,)43( 222 =+ y(1) = 1 and y(1) = 0

Solution: Given222 43 xyxDDx =+ (1)

Take xzorex z log)(, ==

=xD

( )122 = Dx

(1): ( ) zey 22 44 =+ A.E : 2,2,0442 ==+ mmm

C.F = ( ) ( ) 22 log xxBAeBzA z +=+

P.I =( )

( )( )

)1(22

1

2

12

22

2+

=

zz ee

= ( )11

2

2

ze

P.I =

2

22 ze z


23/26

23

=( )

2

log22 xx

(2)

Complete solution is : y = ( ) 2log xxBA + +( )

2

log22 xx

Apply conditions: y(1) =1,y(1) = 0 in (2)A = 1, B = -2

Complete solution is y = ( ) + 2log21 xx ( )

2

log22 xx

EQUATION REDUCIBLE TO THE HOMOGENEOUS LINEAR

FORM (LEGENDRE LINEAR EQUATION)

It is of the form:

( ) ( ) )(...........1

1

11 xfyA

dxydbxaA

dxydbxa nn

n

nn

n

n =+++++

..(

1)

nAAA ....,..........,.........21 , are some constants

It can be reduced to linear differential equation with constant

Coefficients,

by taking: )log()( bxazorebxa z +==+

Consider givesdz

dD

dx

d,, ==

( ) ( ) ( )ybDybxadzdy

bdx

dy

bxa =+=+

Similarly ( ) ( )ybyDbxa 1222 =+ (2)

( ) ( )( )21333 =+ byDbxa and so onSubstitute (2) in (1) gives: the linear differential equation of

constant Coefficients.

Solve ( ) ( ) xyyxyx 612'32''32 2 =++ Solution: This is Legendres linear equation:

( ) ( ) xyyxyx 612'32''32 2 =++ .(1)

Put z = log (2x + 3) , 32 += xez

( ) 232 =+ Dx

( ) ( )dz

dDx ==+ ,432 22

2

Put in (1) : ( ) 931264 2 = zey R.H.S =0

A.E : 01264 2 = mm

4

573,

4

57321

=

+= mm

C.F = zmzm BeAe 21 + C.F = 21 )32()34(

mmxBxA +++


24/26

24

P.I 1 =1264

32

ze= ( )32

14

3+ x

P.I 2 =1264

92

ze

=43

129 =

Solution is

y = ( )( ) ( )

+

+++ 45734/573

3232 xBxA ( )3214

3+ x

4

3

SIMULTANEOUS FIRST ORDER LINEAR EQUATIONS WITH

CONSTANT COEFFICIENTS

1. Solve the simultaneous equations, 023,232 2 =++=++ yx

dt

dyeyx

dt

dx t

Solution: Given 023,232 2 =++=++ yxdt

dyeyx

dt

dx t

Using the operator D =dt

d

( ) teyxD 2232 =++ (1)

( ) 023 =++ yDx .(2)Solving (1) and (2) eliminate (x) :

( ) )3....(..........654)2()2()1(3 22 teyDDD =++ A.E : 0542 =+ mm

m = 1,-5

C.F =tt BeAe 5+

P.I =t

t

eDD

e 22

2

7

6

54

6=

+

y =tt BeAe 5+

te2

7

6

put in (1) : x = ( )[ ]yD 23

1+

x =ttt eBeAe 25

7

8++

solution is :

x = ttt eBeAe 25

7

8++

y =tt BeAe 5+

te2

7

6

2. Solve giventhattxdt

dyty

dt

dx,cos,sin =+=+ t=0, x = 1, y =0

Solution: Dx + y = sin t ..(1)

x + Dy = cost (2)Eliminate x : (1) (2)D ttyDy sinsin2 +=


25/26

25

)3.....(..........sin212 tyD =

1,012 == mm

C.F = Aett Be+

P.I = tt

D

tsin

11

sin)2(

1

sin2

2

=

=

y = Aett Be+ + sint

(2) : x = cost D(y)

x = cost - ( )tBeAedt

d tt sin++

x = tBeAet tt coscos +

x =tt BeAe +

Now using the conditions given, we can find A and B

BAxt +=== 11,0

BAyt +=== 00,0

B =2

1, A =

2

1

Solution is

x = tee tt cosh2

1

2

1=+

y = tttee tt sinhsinsin2

1

2

1=++

3. Solve txdt

dy

tydt

dx

cos2,sin2 ==+

Solution: Dx + 2y = -sin t ..(1)

- 2x +Dy = cos t ...(2)

(1) )(cossin24)2(2 2 tDtyDyD +=++

( ) tD sin342 =+ 2,042 imm ==+

C.F = tBtA 2sin2cos +

P.I = tt

D

tsin

41

sin3

4

sin32

=+

=

+

y = tBtA 2sin2cos + - sin t

(2) : x = [ ]tDy cos2

1

x = ( )

+ tttBtA

dt

dcossin2sin2cos

2

1

x = A cos2t +Bsin2t cost

Solution is :

x = A cos2t +Bsin2t cost

y = tBtA 2sin2cos + - sin t


26/26

26

4. Solve txdt

dy

dt

dxty

dt

dy

dt

dx2sin2,2cos2 =+=+

Solution: Dx + (-D +2)y = cos 2t ..(1)

(D 2)x +Dy = sin 2t (2)

Eliminating y from (1) and (2)

( ) ttxDD 2cos2sin2222

+=+

R.H.S = 0 0222

=+ mm m = i1

C.F = ( )tBtAe t sincos +

P.I1 =( )

D

t

DD

t

+=

+

1

2sin

22

2sin22

=( )

41

)2(sin2sin2sin

1

12

+

=

tDtt

D

D

=

5

2cos22sin tt

P.I 2 = ( )tDD

2cos22

12

+

=( )

10

2sin22cos tt+

x = ++ )sincos( tBtAe t 5

2cos22sin tt ( )

10

2sin22cos tt+

(1) +(2) ttxydt

dx2sin2cos222 +=+

2y = cos2t +sin 2t + 2x -dt

dx2

y =

++

dt

dxxtt 222sin2cos

2

1(3)

Substitute x in (3)

y = ( )2

2sinsincos

ttBtAe t

Solution is :

x = ++ )sincos( tBtAet

5

2cos22sin tt ( )

10

2sin22cos tt+

y = ( )2

2sinsincos

ttBtAe t

unit1 maths

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