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UNIT-I ORDINARY DIFFERENTIAL EQUATIONS - Qualify Gate Exam...

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1 UNIT-I ORDINARY DIFFERENTIAL EQUATIONS Higher order differential equations with constant coefficients – Method of variation of parameters – Cauchy’s and Legendre’s 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, called modeling. (ii) Solutions of this differential equation, evaluating the arbitrary constants from 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 X y K dx y d K dx y d K dx y d n n n n n n n = + + + + - - - - .... .......... 2 2 2 1 1 1 ………….. (1) Where K n K K ..... .......... ,......... 2 1 , are constants. The symbol D stands for the operation of differential (i.e.,) Dy = , dx dy similarly D ... , , 3 3 3 2 2 2 etc dx y d y D dx d y y = = The equation (1) above can be written in the symbolic form (D X y K D K n n n = + + + - ) .......... 1 1 i.e., f(D)y = X Where f (D) = D n n n K D K + + + - . .......... 1 1 Note 1. = Xdx X D 1 2. dx Xe e X a D ax ax - = - 1 3. dx Xe e X a D ax ax - = + 1 (i) The general form of the differential equation of second order is
Transcript

1

UNIT-I

ORDINARY DIFFERENTIAL EQUATIONS

Higher order differential equations with constant coefficients – Method of variation of

parameters – Cauchy’s and Legendre’s 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, called

modeling.

(ii) Solutions of this differential equation, evaluating the arbitrary constants

from 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

ydnn

n

n

n

n

n

=++++ −

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

2

21

1

1 ………….. (1)

Where KnKK ...............,.........21 , are constants.

The symbol D stands for the operation of differential

(i.e.,) Dy = ,dx

dy similarly D ...,,

3

33

2

22 etc

dx

ydyD

dx

dy

y

==

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

(D XyKDK n

nn =+++ − )..........1

1 i.e., f(D)y = X

Where f (D) = D n

nn KDK +++ − ...........1

1

Note

1. ∫= XdxXD

1

2. dxXeeXaD

axax

∫−=

−1

3. dxXeeXaD

axax

∫−=

+1

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

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

dy D

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 form

RQyPDyyD =++2 (Or) ( RyQPDD =++ )2

(iv) Complete solution = Complementary function + Particular Integral

PROBLEMS

1. Solve (D 0)652 =+− yD

Solution: Given (D 0)652 =+− yD

The auxiliary equation is m `0652 =+− m

i.e., m = 2,3

xx BeAeFC 32. +=∴

The general solution is given by

xx 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

3

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

364

2

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 + + xe2

9

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.I xeDD 23

121 +−

=

= xe431

1

+−

= xe0

1

=x xeD 32

1

= x xe32

1

= xxe−

4

P.I ( )xeDD

2

2223

1 −+−

=

= - xe2

264

1

+−

= - x xeD

2

32

1

= - x xe2

34

1

= - xe x2

∴P.I. = P.I 21 .IP+

= xxe− + (- xe x2 )

= -x( xe + e x2 )

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±−=−±−2

2

20164

C.F = e ( )xBxAx sincos2 +−

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 D22 aby −

7. Solve xydx

dy

dx

yd3sin23

2

2

=++

5

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 xDsin

2

1

= xD

xsin

1

2

= ∫ xdxx

sin2

P.I =-2

cos xx

6

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.I 1 =

−+

xD

3cos2

1

1

12

= x3cos19

1

2

1

+−−

= x3cos16

1

P.I 2 =

+x

Dcos

2

1

1

12

= xcos11

1

2

1

+−

= xD

x cos2

1

2

1

= ∫ xdxx

cos4

= xxsin

4

∴P.I = x3cos16

1 + x

xsin

4

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

10. Solve (D xeyD x 2cos)44 22 +=+−

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

The Auxiliary equation is m 0442 =+− m

(m – 2 ) 2 = 0

m = 2 ,2

C.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

0

1

7

= x xe22

2

1

P.I 2 = 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 −

= xD 21

1

−−

= [ ] xD121

−−−

= ( )[ ]xDD .........122 +++−

= [ ]...........000 +++− x

= - x

12. Solve: ( ) 3234 2 xyDDD =+−

8

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 = 1,1

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

P.I = 3

234 2

1x

DDD +−

= ( )[ ]3

22 21

1x

DDD −+

= ( )[ ] 312

221

1xDD

D

−−+

= ( ) ( ) ( )[ ]...............22211 32222

2+−−−+−− DDDDDD

D

= [ ] 3432

24321

1xDDDD

D++++

= [ ]241861 23

2+++ xxx

D

=

+++ x

xxx

D24

2

18

3

6

4

1 334

= 2

24

6

18

12

6

20

2345 xxxx +++

= 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 ( xeyDD x 2cos)522 =+−

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

P.I = xeDD

x 2cos52

12 +−

= ( ) ( )( )1Re2cos

5121

12

=

++−+placeDbyDx

DDe x

9

= xDDD

e x 2cos52212

12

+−−++

= xD

e x 2cos4

12

+

= xe x 2cos44

1

+−

= xD

xe x 2cos2

1

= ∫ xdxxe x

2cos2

P.I = 4

2sin xxe x

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 2

2

+

= e ( ) xD

x sin22

12

2

+−−

= e xD

x sin12

2−

= e xx sin1

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 '

[ ] VDf

DfV

DfxxV

Df 2

'

)(

)(

)(

1

)(

1 −=

10

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

++=

= ( ) ( )[ ] x

DDsin

3141

12 +−+−

= e ( ) xDD

x sin2

12 +

= e( )

xD

Dx sin41

212+−

+−

= [ ]xxe x

sincos25

+−

P.I ( ) xDDe x

3)3(4)3

12

3

2 ++++=

= e ( ) xDD

x

2410

12

3

++

= xDDe x

123

24

101

24

++

= xDe x

−12

51

24

3

=

−12

5

24

3

xe x

General solution is IPFCy .. +=

y = Axx Bee 3−− + [ ]xx

e x

sincos25

+−−

+

−12

5

24

3

xe x

16 Solve xxydx

dy

dx

ydcos2

2

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 +

++−

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 = xD

2

2sin1

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 + x2cos6

1

2

1 +

18. Solve ( ) xexxxydx

yd ++=− 1sin2

2

Solution: A.E : m 012 =−

m = 1±

C.F = A xx Bee +−

P.I ( ) )()(

1

)(

)('

)(

11 V

DfDf

DfxxV

Df

−==

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

22 11

1 +−

=

= ( )[ ] )1(

11

1 2

2x

De x +

−+

= e ( ) )1(2

1 2

2x

DD

x ++

= ( )

12

932 23 xxxe x +−

∴y = A xx Bee +− +( )

12

932 23 xxxe x +−

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 ( ) ( )

−+−

+x

D

Dx

DDxx sin

12

22sin

2

122

sin ,sin xx =α 1,1 22 −=−== αα putD

= e( )( )

−+−

−x

D

Dx

Dxx sin

12

22sin12

12

= e( )( )

( )( )

+−+−

−+−

144

sin22sin

41

2122 DD

xDx

D

Dxx

Put D 12 −=

= e( ) ( )( )

−−+++− xD

DDx

Dxx sin

169

4322sin

5

212

13

= e [ ]

−+−−++−

xD

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 order

equation.

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 = ∫ −− dx

ffff

Xf

2

'

1

1

21

2

Q = ∫ −dx

ffff

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 22

2

'

1

'

21 +=−

= 2 [ ]xx 2sin2cos 22 +

= 2 [ ]1 = 2

P = ∫ −− dx

ffff

Xf

2

'

1

1

21

2

14

= ∫− dxxx

2

2sec2sin

= dxx

x2cos

12sin

2

1∫−

= ∫−

dxx

x

2cos

2sin2

4

1

= )2log(cos4

1x

Q = ∫ −dx

ffff

Xf

2

'

1

'

21

1

= ∫ dxxx

2

2sec2cos

= dxx

x2cos

12cos

2

1∫

= ∫ dx2

1

= x2

1

P.I = Pf 21 Qf+

= )2log(cos4

1x (cos2x) + x

2

1sin2x

2. Solve by the method of variation of parameters xxydx

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 22

2

'

1

'

21 =+=− xxffff

P = ∫ −− dx

ffff

Xf

2

'

1

1

21

2

= dxxxx

∫−1

)sin(sin

= ∫− xdxx 2sin

= ( )

∫−− dx

xx

2

2cos1

= ( )∫ −− dxxxx 2cos2

1

= ∫∫ +− xdxxxdx 2cos2

1

2

1

15

=

−−

+

4

2cos)1(

2

sin

2

1

22

1 2 xxx

x

= xxxx

2cos8

12sin

44

2

++−

Q = ∫ −dx

ffff

Xf

2

'

1

'

21

1

= ∫ dxxxx

1

)(sin)(cos

= ∫ xdxxx cossin

= ∫ dxx

x2

2sin

= ∫ xdxx 2sin2

1

=

−−

−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

( ) 022 =−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 exeeexffff 22222'

12

'

21 22 +−=−

= 2x ( ) ( ) ( )222222 2 xxx eexe −−

= ( ) [ ]12222 −− xxe x

= ( )22xe−

= xe4−

P= ∫ −− dx

ffff

Xf

2

'

1

1

21

2

= ∫ −− dxe

eex

xx

4

22

16

= xdx =∫

Q = ∫ −dx

ffff

Xf

2

'

1

'

21

1

= ∫ −dx

e

exex

xx

4

22

= ∫− xdx

= 2

2x−

P.I = xxx ex

ex

ex 22

22

22

22=−

y = C.F + P.I

= (Ax +B) e x2 + xex 22

2

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

Solution: Given ( ) xyD sec12 =+

The A.E is 012 =+m

m = i±

C.F = xcxc sincos 21 +

= 2211 fcfc +

xfxf sin,cos 21 ==

xfxf cos,sin '

2

'

1 −=

1sincos 22

2

'

1

'

21 =+=− xxffff

P= ∫ −− dx

ffff

Xf

2

'

1

1

21

2

= dxxx

∫−1

secsin

= ∫− dxx

x

cos

sin

= ∫− xdxtan

= log (cos x)

Q = ∫ −dx

ffff

Xf

2

'

1

'

21

1

= dxxx

∫ 1

seccos

= ∫ dx = x

P.I = 21 QfPf +

= xxxx sincos)log(cos +

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

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(cos 22 axaxa +

= a

P.I = 21 QfPf +

P= ∫ −− dx

ffff

Xf

2

'

1

1

21

2

= dxa

axax∫− tansin

= ∫− dxax

ax

a cos

sin1 2

= ∫−−

dxax

ax

a cos

cos11 2

= ( )∫ −−dxaxax

acossec

1

= ( )

−+−a

axaxax

aa

sintanseclog

11

= ( )[ ]axaxaxa

sintanseclog12

−+−

= ( )[ ]axaxaxa

tanseclogsin12

+−

Q = ∫ −dx

ffff

Xf

2

'

1

'

21

1

= ∫ dxa

axax tancos

= ∫ axdxa

sin1

= axa

cos12

21. QfPfIP +=∴

= ( )[ ] [ ]axaxa

axaxaxaxa

cossin1

tanseclogsincos1

22−+−

18

= ( )[ ]axaxaxa

tanseclogcos12

+−

IPFCy .. +=

= axcaxc sincos 21 + ( )[ ]axaxaxa

tanseclogcos12

+−

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= ∫ −− dx

ffff

Xf

2

'

1

1

21

2

= ∫− dxxx

1

tansin

= ∫− dxx

x

cos

sin 2

= dxx

x∫

−−cos

cos1 2

= ( )dxxx∫ −− cossec

= xxx sin)tanlog(sec ++−

Q = ∫ −dx

ffff

Xf

2

'

1

'

21

1

= dxxx

∫ 1

tancos

= ∫ dxsin

= xcos−

21. QfPfIP +=∴

= )tanlog(seccos xxx +−

IPFCy .. +=

= xcxc sincos 21 + )tanlog(seccos xxx +−

8. Solve by method of variation of parameters 144 2

2

''' +=+− xyx

yx

y

Solution: Given 144 2

2

''' +=+− xyx

yx

y

i.e., 24'''2 44 xxyxyyx +=+−

i.e., [ ] 2422 44 xxyxDDx +=+− ………….(1)

Put x = ze

19

Logx = log ze

= z

So that XD = D '

( )1''2 −= DDDx

(1) ( )[ ] ( ) ( )24''' 441 zz eeyDDD +=+−−⇒

[ ] z

eeyDD z 22 4'' 45 +=+−

A.E is 0452 =+− mm

( )( ) 014 =−− mm

4,1=m

zz ececFC 2

4

1. +=∴

Here zz efef == 2

4

1 ,

zz efef == '

2

4'

1 ,

zzz eeeffff 555'

12

'

21 34 −=−=−

21. QfPfIP +=

P= ∫ −− dx

ffff

Xf

2

'

1

1

21

2

= [ ]

∫ −+− dxe

eeez

zzz

5

24

3

= ∫+

dxe

eez

zz

5

35

3

= [ ]dze z

∫−+ 21

3

1

=

−+

23

1 2 zez

= zez 2

6

1

3

1 −−

Q = ∫ −dz

ffff

Zf

2

'

1

'

21

1

= ( )

∫ −+

dze

eeez

zzz

5

344

3

= ∫+− dze

eez

zz

5

68

3

1

= ( )∫ +− dzee zz3

3

1

=

+− z

z

ee

33

1 3

= zz ee3

1

9

1 3 −−

zzzzz eeeeezIP

−−+

−=∴ −

3

1

9

1

6

1

3

1. 342

20

= zzzz eeeze 2424

3

1

9

1

6

1

3

1 −−−

IPFCy .. +=

= zz ecec 2

4

1 + + zzzz eeeze 2424

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

(CAUCHY’S HOMOGENEOUS LINEAR EQUATION)

Consider homogeneous linear differential equation as:

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

11

1 Xyadx

ydxa

dx

yda nx

nn

n

nn =+++ −

−−

(Here a’s are constants and X be a function of X) is called Cauchy’s

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 Euler’s 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 θ,

θ===dz

dxD

dx

dx

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

xdz

dy

xdx

yd +−=

dz

dy

dz

yd

dx

ydx −=∴

2

2

2

22

21

( ) ( )1222 −=−=∴ θθθθDx

Similarly,

( )( )2133 −−= θθθDx

( )( )( )32144 −−−= θθθθDx

and soon.

θ=∴ xD

( )122 −= θθDx

( )( )2133 −−= θθθDx

and so on.

1. Solve )sin(log42

22 xy

dx

dyx

dx

ydx =++

Solution: Consider the transformation

xzorex z log)(, ==

θ=∴ xD

( )122 −= θθDx

( ) )sin(log4122 xyxDDx =++

( ) ( )zy sin412 =+θ

R.H.S = 0 : ( ) 012 =+ yθ

A.E : imm ±==+ ,012

C.F = A cosz + B sinz

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

P.I = ( )zsin41

12 +θ

= 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)(, ==

θ=∴ xD

( )122 −= θθDx

(1) : ( )( ) zey z=++− 241 θθθ

22

( ) zzey =++ 232 θθ

A.E : 0232 =++ mm

M = - 2, -1

C.F = 12 loglog2 −−

+=+ −− xxzz BeAeBeAe

C.F = x

B

x

A +2

P.I = ( )( )ze z

23

12 ++ θθ

=e( ) ( )

zz

2131

12 ++++ θθ

= ( ) ze z

65

12 ++ θθ

= ze z

12

6

51

6

++ θθ

=

−=

−6

5

66

51

6z

ez

e zz

θ

=

−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: Given ( ) 222 43 xyxDDx =+− …………………(1)

Take xzorex z log)(, ==

θ=∴ xD

( )122 −= θθDx

(1) : ( ) zey 22 44 =+− θθ

A.E : 2,2,0442 ==+− mmm

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

P.I = ( )

( )( )

)1(22

1

2

12

22

2 −+=

− θθzz ee

= ( )112

2

θze

P.I = 2

22 z

e z

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

11

1 xfyAdx

ydbxaA

dx

ydbxa nn

nn

n

nn =+++++ −

−−

…………..(

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,, θ==

( ) ( ) ( )ybDybxadz

dyb

dx

dybxa θ=+⇒=+

Similarly ( ) ( )ybyDbxa 1222 −=+ θθ …………(2)

( ) ( )( )21333 −−=+ θθθbyDbxa and so on

Substitute (2) in (1) gives: the linear differential equation of

constant Coefficients.

Solve ( ) ( ) xyyxyx 612'32''322 =−+−+

Solution: This is Legendre’s linear equation:

( ) ( ) xyyxyx 612'32''322 =−+−+ ……………….(1)

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

( ) θ232 =+ Dx

( ) ( )dz

dDx =−=+ θθθ ,432 222

Put in (1) : ( ) 931264 2 −=−− zeyθθ

R.H.S =0

A.E : 01264 2 =−− mm

4

573,

4

57321

−=+= mm

C.F =zmzm

BeAe 21 +

C.F = 21 )32()34(mm

xBxA +++

24

P.I 1=1264

32 −− θθ

ze= ( )32

14

3 +− x

P.I 2 = 1264

92 −− θθ

θze

= 4

3

12

9 −=−

Solution is

y = ( )( ) ( )

−+ +++ 4

5734/5733232 xBxA ( )32

14

3 +− x4

3−

SIMULTANEOUS FIRST ORDER LINEAR EQUATIONS WITH

CONSTANT COEFFICIENTS

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

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 2

2

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

( ) )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

dyty

dt

dxcos2,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 = ( )

−−+ tttBtAdt

dcossin2sin2cos

2

1

x = A cos2t +Bsin2t – cost

Solution is :

x = A cos2t +Bsin2t – cost

y = tBtA 2sin2cos + - sin t

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 = i±1

C.F = ( )tBtAe t sincos +

P.I 1=( )

D

t

DD

t

+=

+−−

1

2sin

22

2sin22

= ( )

41

)2(sin2sin2sin

1

12 +

−=−− tDt

tD

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( tBtAe t

5

2cos22sin tt − ( )10

2sin22cos tt +−

y = ( )2

2sinsincos

ttBtAe t −−


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