ii sem (csvtu) mathematics unit 2 (linear differential equation )solustions

7/28/2019 II sem (csvtu) Mathematics Unit 2 (Linear Differential Equation )Solustions

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epared by Mrityunjoy Dutta

Unit I I Differential Equations of higher order

April -May, 2006

1. Solve: 2 2 32

2 cos2x xd y

y x e e xdx

Ans: 2 2 32 cos2x xD y x e e x ----------- (1)Here Auxiliary equation is 2 2 0 2D D i

So, cos 2 sin 2hy A x B x

2 321

. . cos22

x xpP I y x e e x

D

3 22 21 1

cos2( 3) 2 ( 1) 2

x xpy e x e x

D D

32

22

1 1

cos211 2 361

11

xx

p

e

y x e xD DD D

3

2

2 2

1 1cos2

611 4 2 31

11

xx

p

ey x e x

D D D

2 32 23 22

2

6 66 2 131 ... cos2

11 11 121 1331 4 169

xx

p

D D D De D D Dy x e x

D

3

2 2

2

6 1 36 2 131 ... cos2

11 11 11 121 4 169

xx

p

e D Dy D x e x

D

3

2 2

2

6 47 2 131 ... cos2

11 11 121 4 4 169

xx

p

e D Dy D x e x

3

2 6 47 2 132 2 0 0 ... cos211 11 121 233

xx

p

e Dy x x e x

3

2 12 94 4sin2 13cos211 11 121 233

x x

p

e x ey x x x

So, Solution h py y y

3

2 12 94cos 2 sin 2 4sin2 13cos211 11 121 233

x xe x ey A x B x x x x


2/31


2. Solve: 3 23 23 2

12 2 10( )

d y d yx x y x

dx dx x

Ans:3 2

3 2

3 2

12 2 10

d y d yx x y x

dx dx x

------------- (1)

Let logzx e z x

So,2 3

2 3

2 3', '( ' 1) , '( ' 1)( ' 2)

dy d y d yx D x D D y x D D D y

dx dx dx where

'd

Ddz

Putting all these values in (1) we get

'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e

3 2' ' 2 10( )z zD D y e e -------(2)

Its auxiliary equation is 3 2 2 0m m

2( 1)( 2 2) 0m m m

2 41,

2m

1, 1m i

So, . . cos sin )z zhC F y Ae e B z C z

. . cos(log ) sin(log )hA

C F y x B x C xx

-------------(3)

Now,3 2

1. . 10( )

' ' 2

z zpP I y e e

D D

3 2 3 2

1 1

10 10' ' 2 ' ' 2

z z

py e eD D D D

3 2 3 2

1 110 10 (1)

1 1 2 ( ' 1) ( ' 1) 2

z zpy e e

D D

2

15 10 (1)

'( ' 4 ' 5)

z zpy e e

D D D


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1

21 4 15 10 1 ' ' (1)5 ' 5 5

z zpy e e D D

D

1 45 2 1 ' ..... (1)

' 5

z zpy e e D

D

15 2 (1)'

z zpy e e

D

5 2z zpy e ze

2log5p

xy x

x


2log

cos(log ) sin(log ) 5A x

y x B x C x xx x

(Ans)

3. Solve the following simultaneous equation: 2 , 2 , 2dx dy dzy z xdt dt dt

(Ans.) The given equation is:

2dx

ydt

..(1), 2 ..........(2)dy

zdt

,

2 ..........(3)dz

xdt

Differentiating (1)w . r. t., we get

2

22 2(2 )

d x dyz

dt dt using(2)

Differentiating again w. r. t. t,, we get

3

34 4(2 )

d x dzx

dt dt

3

3 2

21 2 3

2

1 2 3 2 3

2

1 2 3 3

2

1 2

( 8) 0,

8 0 ( 2)() 2 4) 0

2, 1 3

cos( 3 )

1

2

12 cos( 3 ) 3 sin( 3 )

2

2 2cos cos( 3 ) sin sin( 3 )

3 3

cos 3

t t

t t t

t t

t t

dD x whereDdt

D or D D D

D i

x ce c e t c

dxy

dt

ce c e t c c e t c

ce c e t c t c

ce c e t

32

3c


4/31


From(2)..2

1 2 3 2 3

21 2 3

1

2

1 2 22 cos( 3 ) 3 sin( 3 )

2 3 3

4cos( 3 )3

t t t

t t

dyz

dt

ce c e t c c e t c

ce c e t c

Nov-Dec 2006

4. Solve the following : sin , cosdx dxy t x tdt dt

, Given that , 2, 0, , 0x y when t

Ans: Given simultaneous differential equation are

sindx

y tdt

------------- (1)

cosdy

x tdt

------------(2)

From equation (2) we get cosdy

x tdt

---------(3)

Differentiating (3) with respect to t we get2

2sin

dx d yt

dt dt ---------(4)

From (1) and (4) we get2

2sin sin

d yt y t

dt

2

2 2sin

d y

y tdt 2( 1) 2sinD y t -----------(5)

Its Auxiliary equation is 2 1 0 1,1m m

So, . . t thC F y Ae Be

And2

1. . ( 2sin )

1pP I y t

D

2

1( 2sin ) sin

1 1py t t

So, h py y y

sint ty Ae Be t -------(6)

Putting the value of y in (2) we get


5/31


cos cost tAe Be t x t

t tx Ae Be ---------(7)

Given that 2, 0, 0x y when t

So, equation (6) and (7) becomes 0 A B and 2 A B

1, 1A B

So, solutiont tx e e and sint ty e e t (Ans)

5. Solve the differential equation: 3 23 23 2

12 2 10( )

d y d yx x y x

dx dx x

Ans:3 2

3 2

3 2

12 2 10

d y d yx x y x

dx dx x

------------- (1)

Let logzx e z x

So,2 3

2 3

2 3', '( ' 1) , '( ' 1)( ' 2)

dy d y d yx D x D D y x D D D y

dx dx dx where '

dD

dz

Putting all these values in (1) we get

'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e

3 2' ' 2 10( )z zD D y e e -------(2)

Its auxiliary equation is3 2 2 0m m

2( 1)( 2 2) 0m m m

2 41,

2m

1, 1m i

So, . . cos sin )z zhC F y Ae e B z C z

. . cos(log ) sin(log )hA

C F y x B x C xx

-------------(3)


6/31


Now,3 2

1. . 10( )

' ' 2

z zpP I y e e

D D

3 2 3 2

1 110 10

' ' 2 ' ' 2

z zpy e e

D D D D

3 2 3 2

1 110 10 (1)1 1 2 ( ' 1) ( ' 1) 2

z zpy e e D D

2

15 10 (1)

'( ' 4 ' 5)

z zpy e e

D D D

1

21 4 15 10 1 ' ' (1)5 ' 5 5

z zpy e e D D

D

1 45 2 1 ' ..... (1)

' 5

z zpy e e D

D

15 2 (1)

'

z zpy e e

D

5 2z zpy e ze

2log5p

xy x

x


2log

cos(log ) sin(log ) 5A x

y x B x C x xx x

(Ans)

6. Using method of variation of parameters: 22

4 tan2d y

y xdx

Ans: Homogeneous equation is 4 0y y

Its characteristics equation is2 4 0 2i

Hence, Homogeneous solution is cos2 sin2hy A x B x

1 2cos2 , sin2y x y x

1 1 2 2

2 2

cos2 sin22cos 2 2sin 2 2

2sin2 2cos2

y y x xW x x

x xy y

rdxy

y

W

rdxyyyp

.. 12

21

sin2 .tan2 cos2 tan2cos2 sin2

2 2p

x xdx x xdxy x x

cos2 sin2(sec2 cos2 ) sin2 .

2 2p

x xy x x dx xdx

cos2 log(sec2 tan2 ) sin2 sin2 cos2

2 2 2 2 2p

x x x x x xy


7/31


cos2 .log(sec2 tan2 )

4p

x x xy

So, Solutioncos2 .log(sec2 tan2 )

cos2 sin24

h p

x x xy y y A x B x

(Ans)

April -May, 20077. Solve the differential equation 2

22 sinx

d y dyy xe x

dx dx .

Ans:2

22 sinx

d y dyy xe x

dx dx

Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m

So, . . x xhC F y Ae Bxe

Now,2

1. . sin

( 1)

xpP I y xe x

D

2 21 1sin sin

( 1 1)

x xpy e x x e x x

D D

1sinxpy e x xdx

D

1

cos sinxpy e x x xD

cos sinxpy e x x x dx

sin cos cosxpy e x x x x

sin 2cosx

py e x x x So, general solution h py y y

sin 2cosx x xy Ae Bxe e x x x (Ans)

8. Solve by method of variation of parameters of : - 22

sind y

y x xdx

.

Ans: Homogeneous equation is 0y y

Its characteristics equation is2 1 0

2i

Hence, Homogeneous solution is cos sinhy A x B x


1 1 2 2

2 2

cos sincos sin 1

sin cos

y y x xW x x

x xy y


8/31


Wrdxy

yW

rdxyyyp

.. 12

21

sin . sin cos . sincos sin

1 1p

xx xdx xx xdxy x x

2

cos sin . sin sin cospy x x xdx x x x xdx cos sin(1 cos2 ) sin2 .

2 2p

x xy x x dx x xdx

2cos sin2 cos2 sin cos2 sin2

2 2 2 4 2 2 4p

x x x x x x x x xy

2 cos cos sin2 cos cos2 sin cos2 sin sin2

4 4 8 4 8p

x x x x x x x x x x x xy

2 cos sin2 sin cos2cos cos cos2 sin sin24 4 8

p

x x x x xx x x x x xy

2 cos sin cos

4 4 8p

x x x x xy

So, Solution2 cos sin cos

cos sin4 4 8

h p

x x x x xy y y A x B x (Ans)

9. Solve the differential equation 22 22

3 4 (1 )d y dy

x x y xdx dx

.

Ans:2

2 2

23 4 (1 )

d y dyx x y x

dx dx --------- (1)

Let logzx e z x and

2

22' , '( ' 1)

dy d yx D y x D D ydx dx where ' dD dz

Putting all the values in (1) we get

2'( ' 1) 3 ' 4 (1 )zD D D y e 2 2' 4 ' 4 1 2 z zD D y e e ----------- (2)Its auxiliary equation is 2 24 4 0 ( 2) 0 2,2m m m m

So, 2 2 2 2. . logz zhC F y Ae BZe Ax Bx x

Now,2

2

1. . (1 2 )

( ' 2)

z zpP I y e e

D

0 2

2 2 2

1 1 1

2( ' 2) ( ' 2) ( ' 2)

z z z

py e e eD D D

2

2 2 2

1 1 1.1 2 1

(0 2) (1 2) ( ' 2 2)

z zpy e e

D

2

2

1 12 1

4 '

z zpy e e

D 2

12 1

4

z zpy e e dzdz


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2 2 2 21 1 (log )2 2

4 2 4 2

zz

p

z e x xy e x

So, general solution is2 2

2 2 1 (log )log 24 2

h p

x xy y y Ax Bx x x (Ans).

Nov-Dec 2007

10.Solve: 22

3 4 0d y dy

ydx dx

Sol. Its symbolic form is:2

2

3 4 0

( 3 4) 0

D y Dy y

D D y

Its auxiliary equation is :2 3 4 0D D

4, 1D

Hence C.F.=4

1 2x xy ce c e

(Ans.)

11.Solve: 22

5 6 sin3d y dy

y xdx dx

Sol. It s symbolic form is 2 5 6 tan2D D y x

Its auxiliary equation is: 2 5 6 0D D ( 2)( 3) 0

2,3

D D

D

Hence C.F.= y=2 3

1 2x xce c e

And P.I.=

2 2

2 2

1 1sin3 sin3

( 5 6) 3 5 6

1 1 5 3sin3 [ ][ ]sin3

5 3 5 3 5 3

5 3 5 3sin3 sin3

25 9 25( 3 ) 9

1 15cos3 3sin3(5 3)sin3234 234

5cos3 sin3

78

x xD D D

Dx x

D D D

D Dx x

D

x xD x

x x

Hence complete solution is:y=C.F.+P.I.

2 31 2

1(5cos3 sin3 )

78

x xy ce c e x x


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12.Solve by method variation of parameters:2

24 tan2

d yy x

dx


Its characteristics equation is2 4 0 2i

Hence, Homogeneous solution is cos2 sin2hy A x B x


1 1 2 2

2 2


2sin2 2cos2

y y x xW x x

x xy y

rdxy

yW

rdxyyyp

.. 12

21

sin2 .tan2 cos2 tan2cos2 sin22 2

p x xdx x xdxy x x


2 2p

x xy x x dx xdx


2 2 2 2 2p

x x x x x xy


4p

x x xy


cos2 sin2

4

h p

x x xy y y A x B x

(Ans)

13.Solve the simultaneously equation:2 5 , 4 3t

dx dxx y e x y t

dt dt

Sol. The given equation can be expressed as:

( 5) 3 ............(1)

2 ( 5) ......(2)tD x y t

x D y e

To eliminate y , operating equation (1) by (D+5) and equation (2) by 3 then subtracting , we get

2

( 5)( 4) 6 ( 5) 3

( 9 14) 1 5 3

t

t

D D x x D t e

D D x t e

..(3)

The root of auxiliary equation of the equation corresponding homogeneous equation

2( 9 14) 0D D x

of the equation (3) is given by


11/31


2( 9 14) 0

, 2, 7

D D

or D

Hence the complementary function of equation (3) is:

2 71 2. .

t tC F ce c e

The particular integral of equation (3) is

2

2 2

2

1. . (1 5 3 )

( 9 14)

1 1(1 5 ) 3

( 9 14) ( 9 14)

1 9 31 ............. (1 5 )

14 14 14 1 9 14

1 91 5 .5

14 14 8

1 315

14 14 8

t

t

t

t

t

P I t eD D

t eD D D D

D eD t

et

et

Hence the general solution of equation (3) is:

2 7

1 2

2 71 2

. . . .

1 315

14 14 8

5, 2 7

14 8

tt t

tt t

x C F P I

ex ce c e t

dx eNow ce c e

dt

.(4)

Substituting the above values of x and dx/dt in equation (1), we get;

2 7 2 71 2 1 2

2 71 2

5 20 1243 2 7 4 4

14 8 14 196 2

1 3 5 272 3 .......................(5)

3 7 8 98

t tt t t t

t t t

e ey ce c e ce c e t t

y ce c e t e

Since the degree of D in the determinant4 3

,22 5

Dis

D

it follows that the number of

independent constant in general solution must be two. Hence (4) and (5) together constitute the

general solution of the given system.

April -May, 2008

14.State Cauchys Linear equation.Ans: - A Differential equation of the form


12/31


Xyad

ydxa

d

ydxa

d

ydx nn

nn

n

nn

n

nn

........

2

22

21

11

1 is known as Cauchys

Linear differential equation.

15.Solve2

2

2 4 sin2xd y dy

y e xdx dx .

(ANS). Its symbolic form is :2 2( 4 1) sin2xD D y e x

Its auxiliary equation is:2( 4 1) 0D D

16.Solveby method of variation of parameters of : - 22

4 4tan2 .d y

y xdx


Its characteristics equation is2

4 0 2i Hence, Homogeneous solution is cos2 sin2hy A x B x


1 1 2 2

2 2


2sin2 2cos2

y y x xW x x

x xy y

rdxy

yW

rdxyyyp

.. 12

21

sin2 .tan2 cos2 tan2cos2 sin2

2 2p

x xdx x xdxy x x


2 2p

x xy x x dx xdx


2 2 2 2 2p

x x x x x xy


4p

x x xy


cos2 sin24

h p

x x xy y y A x B x

(Ans)

17.Solve 2 0dydx x ydt dt

and 5 3 0dydx

x ydt dt

(Ans). The given equation are:

2 0.....................(1)dydx

x ydt dt


13/31


5 3 0.....................(2)dydx

x ydt dt

Subtracting equation (2) from equation (1), we get

3 2 0...............(3)dx

x ydt

Writing equation (2)and (3) symbolically (i.e;D=d/dt),we have5 ( 3) 0..........(4)

( 3) 2 0..........(5)

x D y

D x y

Operating equation (5)by D+3 ,we get2( 9) 2( 3) 0.........(6)D x D y

Now, multiplying equation(4) by 2 and adding in equation (6),we get2( 1) 0D x .(7)

This is the linear equation in x with constant coefficient s for which the auxiliary equation is2( 1) 0D

D i

Hence the general solution of linear equation is:0

1 2

1 2

( cos sin )

( cos sin )..........................(8)

tx e c t c t

x c t c t

From equation (8)

1 2

1 2

sin cos

sin cos

dxc t c t

dt

Dx c t c t

Substituting these values in equation(5) we get

1 2 1 2

2 1 1 2

1 1( 3 ) ( sin cos ) 3( cos sin )

2 2

1 1( 3 )cos ( 3 )sin

2 2

y Dx x c t c t c t c t

y c c t c c t

Nov-Dec 200818.Define the linear differential equation.Ans. A differential equation in which the dependent variable & its derivatives occur only in the first

degree & are not multiplied together is known as Linear Differential equation.

General form of Linear Differential Equation of nth order is:

XyPd

ydP

d

ydP

d

ydnn

n

n

n

n

n

........2

2

21

1

1 , where XPPP n,,........,, 21 are functions of x.

19.Solve the differential equation:


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2

23 2

xd y dy ey edx dx

Ans: - Its symbolic form is 2 3 2xeD D y e .

Its Auxiliary equation is 2 3 2 0D D 1, 2D

Hence, Complementary function is 2x xcy Ae Be

Now, 2 1 1 1 13 2 ( 2)( 1) ( 2) ( 1)x x xe e ePI e e e

D D D D D D

2 2 21 1

2 2

x x x x x x x xx x x xe e e ePI e e e dx e e e e e e dx e e e dxD D

2x xePI e e

So, Solution 2 2x x xc pxey y y Ae Be e e (Ans)

20.Solve by method variation of parameters:2 3

2 26 9

xd y dy ey

dx dx x

Ans: -2

3

2

2

96e

yd

dy

d

yd x

Its Homogeneous equation is 0962

2

yd

dy

d

yd

Its symbolic form is 0)96( 2 yDD

Its characteristics equation is 0)96( 2 DD

3,3 D

Hence, Homogeneous solution is 213)( ByAyeBxAy xh

xx xeyey 323

1 ,

xxxx

xxx

xx

exexeexeee

xee

yy

yyW 6666

333

33

21

2133

33''

Wrdxy

yW

rdxyyyp

.. 12

21

dxee

xedxexe

ey

x

x

xx

x

x

xx

p 2

3

6

33

2

3

6

33

..

dxxedxeyxx

p 2

33 11

xxexey xe

xp

1log 33 xe

xp exey

33 log

So, Solutionx

exxx

ph exeBxeAeyyy3333 log (Ans).


15/31


21.Solve the simultaneously equation:5 2 , 2 0

dx dyx y t x y

dt dt being x=y=0 when t=0.

Ans: - +5 2 = , +2 + = 0Its symbolic form is ( +5) 2 = ------------------(1)

2 +( +1) = 0-------------------(2)Now, eq(1) X (D+1) +eq(2) X 2 we get

[( +1)( +5) +4] = ( +1) +0 ( +6 +9) = 1+, which is linear differential equation.Its auxiliary equation is ( +6 +9) = 0 =3,3So,..= ( +)

Now, ..= () (1+ ) = 1+

(1+) = 1 2 +3

+ . (1+)

..= 1+ = + = + So, = ( +) + + ---------------------------------(3)Now, from (1) 2 = ( +5) = +5 2 = 5( +) + + 3( +) + + = (2 +) + + ---------------(4)Given that at = 0, = = 0, so equation (3) and (4) becomes

+ = 0 = and

(2 +) +

= 0 =

2 ==

+

=

=

So, Solution is

= ( ) + + and = + + +

= {(1+6) (3 +1)} and = {(4+6) +(6 4)}April -May, 2009

22.Explain briefly the method of variation of parameter.Ans: Let us solve

2

2

d y dyP Qy X

dx dx by variation of parameter method.

Let its complementary function is 1 1 2 2cy c y c y

Then find out wronskian1 2

1 2

1 2

( , )' '

y yW y y

y y

Now, particular integral 2 11 2

1 2 1 2

. .

( , ) ( , )p

y X y Xy y dx y dx

W y y W y y

Then general solution is c py y y (Ans).


16/31


23.Solve the differential equation 2 2 32 cos2x xD y x e e x .Ans: 2 2 32 cos2x xD y x e e x ----------- (1)

Here Auxiliary equation is 2 2 0 2D D i

So, cos 2 sin 2hy A x B x

2 321

. . cos22

x xpP I y x e e x

D

3 22 21 1

cos2( 3) 2 ( 1) 2

x xpy e x e x

D D

3

2

2 2

1 1cos2

611 2 31

11

xx

p

ey x e x

D D D D

3

2

2 2

1 1cos2

611 4 2 3111

xx

p

ey x e x

D D D

2 32 23 2

2

2

6 66 2 131 ... cos2

11 11 121 1331 4 169

xx

p

D D D De D D Dy x e x

D

3

2 2

2

6 1 36 2 131 ... cos2

11 11 11 121 4 169

xx

p

e D Dy D x e x

D

32 2

2

6 47 2 131 ... cos2

11 11 121 4 4 169

xx

p

e D Dy D x e x

3

2 6 47 2 132 2 0 0 ... cos211 11 121 233

xx

p

e Dy x x e x

3

2 12 94 4sin2 13cos211 11 121 233

x x

p

e x ey x x x


3

2 12 94cos 2 sin 2 4sin2 13cos211 11 121 233

x xe x ey A x B x x x x

24.Solve the equation : - 22 2(1 ) (1 ) sin 2log(1 )d y dyx x y xdx dx .Ans:

22

2(1 ) (1 ) sin 2log(1 )

d y dyx x y x

dx dx (1) is a Legendres equation.

Now let us put 1 tx e we get2

2

2(1 ) ( 1) , (1 )

d y dyx D D y x Dy

dx dx where

dD

dt


17/31


So, equation (1) becomes ( 1) sin2D D y Dy y t 2( 1) sin2D y t which is linear differential equation with constant coefficients.

Its auxiliary equation is 2( 1) 0D D i

So, 1 2. . cos sinC F c t c t ..(2)

Now,2 2

1 sin2 sin2. . sin2

1 2 1 3

t tP I t

D

So, solution 1 2sin2

. . . . cos sin3

ty C F P I c t c t (Ans)

25.Solve the simultaneous equation: -2 , 2t t

dx dyy e x e

dt dt

.

Ans: 2 , 2t tdx dy

y e x edt dt

2 ...............(1) 2 ..............(2)t tDx y e Dy x e whered

Ddt

(1) 2 (2) D we get22 4 2 2 t tDx y D y Dx e e

2( 4) 2 t tD y e e ..(3)

HereIts auxiliary equation is 2( 4) 0 2D D i

So, 1 2. . cos2 sin2C F c t c t ..(4)

Now, 2 2 21 1 1

. . 2 2

1 1 1

t t t tP I e e e e

D D D

2 2

1 1. . 2

1 1 ( 1) 1 2

tt t t eP I e e e

So, solution 1 2. . . . cos2 sin22

tt ey C F P I c t c t e

..(5)

Now putting the value of y in (2) we get

2 tDy x e

1 22 sin2 2 cos2 22

tt tec t c t e x e

1 22 2 sin2 2 cos22

tt ex c t c t e

1 2sin2 cos22 4

t te ex c t c t

(6)

So, solution 1 2 1 2sin2 cos2 , cos2 sin22 4 2

t t tte e ex c t c t y c t c t e

(Ans)


18/31


Nov-Dec 2009

26.Solve the equation: 2( 2) 0D D y .Ans: - 2( 2) 0D D y

Its auxiliary equation is 2( 2) 0 1, 2D D D

So, solution is 2x xy Ae Be (Ans).

27.Solve the following differential equation:2

2

2log

d y dyx x y x

dx dx .

Ans: - + = --------------------(1)

Given equation is Cauchys Homogeneous Linear differential equation.

Put

=

=

, if

=

,

=

,

=

(

1)

So, equation (1) is ( 1) + = ( 2 +1) = Which is linear differential equation with constant coefficients.

So, its Auxiliary equation is 2 + 1= 0 = 1,1So,..= + ------------------------(2)

Now, ..= () = (1 ) = (1+2 +3 + .)

..= +2So, solution is =..+..= + + +2 = + + +2 (Ans).

OR

28.Solve the simultaneous differential equation:

5 2 , 2 0dx dy

x y t x ydt dt

, given that 0x y when 0t .

Ans: - +5 2 = , +2 + = 0

Its symbolic form is ( +5) 2 = ------------------(1)2 +( +1) = 0-------------------(2)

Now, eq(1) X (D+1) +eq(2) X 2 we get

[( +1)( +5) +4] = ( +1) +0 ( +6 +9) = 1+, which is linear differential equation.Its auxiliary equation is

(

+6

+9) = 0

=

3,

3So, ..= ( +)

Now, ..= () (1+ ) = 1+

(1+) = 1 2 +3

+ . (1+)

..= 1+ = + = + So, = ( +) + + ---------------------------------(3)Now, from (1) 2 = ( +5) = +5


19/31


2 = 5( +) + + 3( +) + + = (2 +) + + ---------------(4)Given that at = 0, = = 0, so equation (3) and (4) becomes

+

= 0

=

and

(2 +) + = 0 = 2 == + = = So, Solution is

= ( ) + + and = + + +

= {(1+6) (3 +1)} and = {(4+6) +(6 4)}29.Solve the following differential equation: 2

24 sinh

d yy x x

dx .

Ans: - 4 =

4 = Its Symbolic form is ( 4) =

Its Auxiliary equation is ( 4) = 0 = 2So, ..= +Now, ..=

=

.

.=

()

()

=

..=

+

..=

1+ + +

1+

+

..=

+ + = (

) (

)

..= [3 2]So, solution is =..+..= + + [3 2] (Ans).

April -May, 201030.Define Cauchy and Legendre L inear differential equation.

Ans: - A Differential equation of the form


20/31


Xyad

ydxa

d

ydxa

d

ydx nn

nn

n

nn

n

nn

........

2

22

21

11



A Differential equation of the form

Xyad

ydbaxa

d

ydbaxa

d

ydbax nn

nn

n

nn

n

nn

........)()()(

2

22

21

11

1 is

known as Legendres Linear differential equation.

31.Solve 33

3 2xd y dy ey e

dx dx .

Ans: - Its symbolic form is 2 3 2xeD D y e .

Its Auxiliary equation is2 3 2 0D D 1, 2D

Hence, Complementary function is2x x

cy Ae Be

Now, 2 1 1 1 13 2 ( 2)( 1) ( 2) ( 1)x x xe e ePI e e e

D D D D D D

2 2 21 1

2 2

x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D

2 xx ePI e e

So, Solution 2 2xx x x e

c py y y Ae Be e e (Ans)

32.Solve, by the method of variation of parameters: xeDD x log)12( 2 .Ans: - Its Auxiliary equation is

2 2 1 0D D 1, 1D

Hence, Complementary function is x

cy A Bx e

1 2,x xy e y xe

1 2 2 2 2 2

1 2

x xx x x x

x x x

y y e xeW e xe xe e

e e xey y

WXdxy

yW

Xdxyyyp

.. 12

21

2 2

. log . logx x x xx xp x x

xe e xdx e e xdxy e xe

e e

log logx xpy e x xdx xe xdx

2 2

log log2 4

x xp

x xy e x xe x x x

2 2 23

log 2log 32 4 4

x x x

p

x x xy e x e e x


21/31


So, Solution 2

( ) 2log 34

x x

c p

xy y y A Bx e e x (Ans)

33.Solve the simultaneous equation: -0 y

d

dxt , 0 x

dt

dyt , given that 0)1(,1)1( yx .

Ans: - 0 yd

dxt ------------------- (1), 0 x

d

dyt ------------- (2)

Differentiating (1) w.e.t. t we get2

20

dx d x dyt

dt dt dt , by multiplying t both side we get

22

20

dx d x dyt t tdt dt dt

-------------------- (3)

By putting (2) in (3) we get2 2

2 2

2 20

d x dx d xt t t x

dt dt dt -----(4) which is Cauchy

Linear Differential Equation.

Put logzt e z t and2

2

2( 1) ,

d x dxt D D x t Dx

dt dt where

dD

dz

So, eqn (4) becomes 2( 1) 0 ( 1) 0D D x Dx x D x

Its Auxiliary equation is 2( 1) 0 1,1D D

So, z zB

x Ae Be Att

-------(5)

Putting in (1) we get2

dx B By t t A At

dt t t

----------- (6)

So, Solution is Bx Att

and By Att

(Ans).

Given x(1) =1 and y(-1) =0, so2

1,

2

10,1 BAABBA

Hence solution is

t

tyBA

ttx

1

2

1,

1

2

1(Ans).

Nov-Dec 201034.Write the formula for P.I. for the method of variation of parameters.

Ans: - Let equation is rybaDD )( 2 , then P.I is given by

Wrdxy

yW

rdxyyyp

.. 12

21 where

'' 21

21

yy

yyW .

35.Solve )2sin(8)2( 222 xxeyD x .Ans: - )2sin(8)2( 222 xxeyD x --------------------- (1)

Its characteristics equation is 2,20)2( 2 DD


22/31


So, xxh BxeAeyFC22.. ----------------------- (2)

Now, )2sin(8)2(

1.. 22

2xxe

DyIP xp

2

22

2

2 )2(

1

2sin)2(

1

)2(

1

8 xDxDeDy

x

p

222

2

21

1

4

12sin

44

1

)2(2

18 x

Dx

DDe

Dxy xp

2

2

2

22 ..........4

32

214

12sin

442

1

2

18 x

DDx

Dexy xp

..........004

624

1

2

2cos

4

1

2822

2

xxx

ex

yx

p

3422cos4 222 xxxexy xp

So, Solution 3422cos4 22222 xxxexBxeAeyyy xxxph (Ans).

36.Solve, by the method of variation of parameters:2

3

2

2

96e

yd

dy

d

yd x .

Ans: -2

3

2

2

96e

yd

dy

d

yd x

Its Homogeneous equation is 0962

2

yd

dy

d

yd

Its symbolic form is 0)96( 2 yDD

Its characteristics equation is 0)96( 2 DD

3,3 D

Hence, Homogeneous solution is 213)( ByAyeBxAy xh

xx xeyey 323

1 ,

xxxx

xxx

xx

exexeexeee

xee

yy

yyW 6666

333

33

21

2133

33''

Wrdxy

yW

rdxyyyp

.. 12

21

dxee

xedxexe

eyx

x

xx

x

x

xx

p 2

3

6

33

2

3

6

33 ..

dxxedxeyxx

p 2

33 11


23/31


xxexey xe

xp

1log 33 xe

xp exey

33 log

So, Solutionx

exxx

ph exeBxeAeyyy3333 log (Ans).

37.Solve: xyd

dy

xd

yd

x elog2

22

.

Ans: - xyd

dyx

d

ydx elog2

22 --------- (1)

Let logzx e z x and2

2

2' , '( ' 1)

dy d yx D y x D D y

dx dx where '

dD

dz

Putting all the values in (1) we get

zyDDzyDDD ]12[1)1( 2 ----------- (2)Its auxiliary equation is 1,10122 mmm

So, xBxAxBzeAeyFC ezz

h log..

Now, zD

yIP p 2)1(

1..

zDDzDyp .........)321()1(22

xzzy ep log22........0002

So, general solution is xxBxAxyyy eeph log2log (Ans).

April -May, 2011

38.Find the particular integral of Solve

3

3. 4 sin2

d y dy

xdx dx

Ans.

3

3.4 sin2

d y dyx

dx dx

P.I will be3

1sin2

4PI x

D D

3 21 1

sin2 sin24 4

x xD D D D

3 2

1 1sin2 sin2

4 2 4

x x

D D D

31 1

sin2 sin24 0

x xD D D

By differentiating3 4D D as

1, 0in f D cannot bef D


24/31


3 21 1

sin2 sin24 3 4

x x xD D D

3 21 1

sin2 sin24 3( 2 ) 4

x x xD D

31 sin2 sin24 8

xx xD D

3

1 sin2sin2

4 8

x xx

D D

P.I of

3

3.4 sin2

d y dyx

dx dx will be

sin2

8

x x

39.Solve 22

3 2xd y dy ex y e

dx dx

Ans: - Its symbolic form is 2

3 2

xe

D D y e .Its Auxiliary equation is

2 3 2 0D D 1, 2D

Hence, Complementary function is2x x

cy Ae Be

Now, 21 1 1 1

3 2 ( 2)( 1) ( 2) ( 1)

x x xe e ePI e e eD D D D D D

2 2 21 1

2 2

x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D

2 xx ePI e e

So, Solution2 2x x

c p

xx ey y y Ae Be e e

(Ans)

40.Solve the differential equation: 222

5 4 .log .d y dy

x x y x xdx dx

(Ans). Let

zx e andd

Ddz

. Then2

2

2, ( 1)

dy d yx Dy x D D y

dx dx

Thus the given differential equation reduces to the following fotm:

2

( 1) 5 4

( 4 4)

z

z

D D D y ze

D D y ze

Which is the linear differential equation with constant coefficient s, for which the auxiliaryequation is:

2 4 4 0D D 2, 2D

C.F= 2 21 2 1 2 logzc c z e c c x x

P.I Will be


25/31


1

( )

zzef D

i.e

2

1.

( 4 4)

zP I zeD D

2 2 21. ( 4 4) ( 1) 4( 1) 4 ( 6 9)

z z

z e eP I ze z zD D D D D D

26. (1 )

9 9 9

ze D DP I z

6.

9 9

zeP I z

2. log

9 3

xP I x

Complete Ans is C.F +P.I

21 2 logc c x x + 2log

9 3x x

41.Solve sin , cos ,dx dyy t x tdt dt

Given that x=2 and y=0 when t=0.

Ans: Given simultaneous differential equation are

sindx

y tdt

-------------(1)

cosdy

x tdt

------------(2)

From equation (2) we get cosdyx tdt

---------(3)

Differentiating (3) with respect to t we get

2

2sin

dx d yt

dt dt

---------(4)

From (1) and (4) we get

2

2sin sin

d yt y t

dt

2

2

2sind y

y tdt

2( 1) 2sinD y t -----------(5)

Its Auxiliary equation is2 1 0 1,1m m

So,. . t thC F y Ae Be


26/31


And2

1. . ( 2sin )

1pP I y t

D

2

1( 2sin ) sin

1 1py t t

So, h py y y

sint ty Ae Be t -------(6)

Putting the value of y in (2) we get

cos cost tAe Be t x t t tx Ae Be ---------(7)

Given that2, 0, 0x y whent

So, equation (6) and (7) becomes 0 A B and 2 A B 1, 1A B

So, solution t tx e e and sint ty e e t (Ans)

Nov-Dec 2011

42. Explain Cauchys homogeneous linear differential equation.Ans: - A Differential equation of the form

Xyad

ydxa

d

ydxa

d

ydx nn

nn

n

nn

n

nn

........

2

22

21

11



43.Solve the differential equation2

22 sinxd y dy y xe x

dx dx .

Ans:2

22 sinx

d y dyy xe x

dx dx

Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m

So, . . x xhC F y Ae Bxe

Now,2

1. . sin

( 1)

xpP I y xe x

D

2 2

1 1sin sin

( 1 1)

x x

p

y e x x e x xD D

1sinxpy e x xdx

D

1

cos sinxpy e x x xD

cos sinxpy e x x x dx


27/31


sin cos cosxpy e x x x x

sin 2cosxpy e x x x

So, general solution h py y y

sin 2cosx x xy Ae Bxe e x x x (Ans)

44.Solve the differential equation 222

log .sin(log ) d y dy

x y x xdx dx

zx e andd

Ddz

. Then2

2

2, ( 1)

dy d yx Dy x D D y

dx dx


2

( 1) 1 (sin )

( 1) sin

D D D y z z

D y z z

Which is the linear differential equation with constant coefficient s, for which the auxiliary

equation is:2 1 0D

,D i i

C.F= 1 2 1 2cos sin cos log sin logc z c z c x c x P.I Will be

1sin

( )z z

f Di.e

2 2

1 1

. sin .( 1) ( 1)

iz

P I z z I P zeD D

22

1. sin .

( 1) 1)

izeP I z z I P z

D D i

21

. sin .( 1) 2

izeP I z z I P z

D D i

2

1 1 1. sin .

( 1) 2

1 2

izP I z z I Pe zDD i

i

1

2

1 1. sin . 1

( 1) 2 2

DizP I z z I Pe zD i i

2

1 1. sin . 1

( 1) 2 2

DizP I z z I Pe zD i i


28/31


2

1 1 1. sin .

( 1) 2 2

izP I z z I Pe zD i i

2

1. sin .

( 1) 2 2

i iizP I z z I Pe zD

21 1. sin .

( 1) 2 4iizP I z z I Pe z

D

21 1

. sin . cos sin( 1) 2 4

iP I z z I P z i z z

D

2

1 cos sin sin. sin . cos

( 1) 2 4 2 4

i z z z i zP I z z I P z z

D

2

1 cos sin sin 1. sin . cos

( 1) 4 2 4 2

z z z zP I z z I P i z z

D

2

1 sin 1

. sin cos( 1) 4 2

z

P I z z z zD

2

1 sinlog 1. sin log coslog

( 1) 4 2

xP I z z x x

D

Ans.= C.F+P.I

1 2cos log sin logc x c x +sinlog 1

log coslog4 2

xx x

45.Solve the following simultaneous equation 2dx ydt

, 2dy

zdt ,

2 .dz

xdt

(Ans.) The given equation is:

2dx ydt

..(1), 2 ..........(2)dy zdt

,

2 ..........(3)dz xdt

Differentiating (1)w . r. t., we get

2

22 2(2 )

d x dyz

dt dt using(2)

Differentiating again w. r. t. t,, we get

3

34 4(2 )

d x dzx

dt dt

3

3 2

2

1 2 3

( 8) 0,

8 0 ( 2)() 2 4) 0

2, 1 3

cos( 3 )

1

2

t t

dD x whereD

dt

D or D D D

D i

x ce c e t c

dxy

dt


29/31


2

1 2 3 2 3

2

1 2 3 3

21 2 3

12 cos( 3 ) 3 sin( 3 )

2

2 2cos cos( 3 ) sin sin( 3 )

3 3

2cos 33

t t t

t t

t t

ce c e t c c e t c

ce c e t c t c

ce c e t c

From(2)..2

1 2 3 2 3

21 2 3

1

2

1 2 22 cos( 3 ) 3 sin( 3 )

2 3 3

4cos( 3 )

3

t t t

t t

dyz

dt

ce c e t c c e t c

ce c e t c

April -May, 2012

46.Solve3

3 0d y ydx

Ans: - Its symbolic form is Solve 013 yD .Its Auxiliary equation is 0)1)(1(01 23 DDDD

2

31,

2

31,1

iiD

So, solution 23

23221 sincos..x

xxx eCCeCyFC (Ans).

47.Solve 222

5 4 logd y dy

x x y x xdx dx

(Ans). Let

zx e andd

Ddz

. Then2

2

2, ( 1)

dy d yx Dy x D D y

dx dx


2

( 1) 5 4

( 4 4)

z

z

D D D y ze

D D y ze

Which is the linear differential equation with constant coefficient s, for which the auxiliary

equation is:2 4 4 0D D

2, 2D

C.F= 2 21 2 1 2 logzc c z e c c x x

P.I Will be

1

( )

zzef D

i.e


30/31


2

1.

( 4 4)

zP I zeD D

2 2 2

1.

( 4 4) ( 1) 4( 1) 4 ( 6 9)

z zz e eP I ze z z

D D D D D D

26. (1 )9 9 9

ze D DP I z

6.

9 9

zeP I z

2. log

9 3

xP I x

Complete Ans is C.F +P.I

21 2 logc c x x +

2log

9 3

xx

48.Solve :

3 2

3. 22 sin2x

d y d y dye x

dx dx dx

Sol. Its symbolic form is xeyDDD x 2sin2 23 Its Auxiliary equation is 0)12(02 223 DDDDDD

1,1,0 D

So, xxc exCCeCyFC 32

01..

xc exCCCy 321 ---------------------(1)

Now, xeDDD

yIP xp 2sin2

1..

23

xDDD

eDDD

y xp 2sin2

1

2

12323

xDDDD

eDD

xy xp 2sin2.

1

13

1222

xDD

eD

xy xp 2sin)4(2)4.(

1

46

12

xD

exy

x

p 2sin83

1

4)1(62

xD

Dexy

x

p 2sin69

83

2 2

2

xDex

yx

p 2sin64)4(9

83

2

2


31/31

10

2sin82cos6

2

2 xxexy

x

p

-------------------(2)

So, solution 10

2sin82cos6

2

2

321

xxexexCCCyyy

xx

pc

(Ans).

49.Solve the equations simultaneously5 2 , 2 0, 0

dx dyx t t x y x y

dt dt when t=0

Ans: - +5 2 = , +2 + = 0

Its symbolic form is ( +5) 2 = ------------------(1)2 +( +1) = 0-------------------(2)

Now, eq(1) X (D+1) +eq(2) X 2 we get

[( +1)( +5) +4] = ( +1) +0 ( +6 +9) = 1+, which is linear differential equation.Its auxiliary equation is ( +6 +9) = 0 =3,3So, ..= ( +)

Now, ..= () (1+ ) = 1+

(1+) = 1 2 +3

+ . (1+)

..= 1+ = + = + So, = ( +) + + ---------------------------------(3)Now, from (1) 2 = ( +5) = +5 2 = 5( +) + + 3( +) + + = (2 +) + + ---------------(4)Given that at = 0, = = 0, so equation (3) and (4) becomes + = 0 = and

(2 +) + = 0 = 2 == + = = So, Solution is

= ( ) + + and = + + +

= {(1+6) (3 +1)} and = {(4+6) +(6 4)}

ii sem (csvtu) mathematics unit 2 (linear differential equation )solustions

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