variation of parameters

7/27/2019 Variation of Parameters

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MATH C241:MATHEMATICSIII

BITS-PILANI HYDERABAD CAMPUS

7

3-Sep-12 1

Presented by

Dr. Akhlad Iqbal


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3-Sep-12 2

Variation of parameters

Ch. 3

George F. Simmons,Differential Equations with

Applications and Historical notes,Tata McGraw-Hill, 2nd Ed, 2003(Twelfth reprint, 2008)

Lecture 7


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3-Sep-12 3

The method of

Variation of Parameters


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3-Sep-12 4

In this lecture we discuss a generalmethod of finding a particular solution of

a non-homogeneous l.d.e. (whether it isconstant coefficient equation or not).


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3-Sep-12 5

Consider the second order l.d.e.

0 1 2( ) ( ) ( ) ( )..(*)a x y a x y a x y h x + + =

We assume that we have already found the

C.F. as

1 1 2 2( ) ( )y c y x c y x= +

The Method of Variation ofparameters


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3-Sep-12 6

The Method of Variation of parameters says:Take a particular solution of (*) as

1 1 2 2( ) ( )y v y x v y x= + ($)

where v1(x), v2(x) are functions ofx to be

chosen such that ($) is a solution of (*).


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3-Sep-12 7

1 1 2 2( ) ( )y v y x v y x= + ($)Differentiating ($) w.r.t.x, we get

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )y v y x v y x v y x v y x = + + +

We now choose v1, v2 such that

1 1 2 2( ) ( ) 0 .....(1)v y x v y x + =

Thus1 1 2 2( ) ( )y v y x v y x = +


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3-Sep-12 8

Differentiatingy w.r.t.x, we get

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )y v y x v y x v y x v y x = + + +

Substituting fory,y,y in (*), we get

[ ][ ]

[ ]

0 1 1 2 2 1 1 2 2

1 1 11 2 22

2 1 1 2 2

( )( )

( ) ( )

a x v y v y v y v ya x v y v y

a x v y v y h x

+ + + +

+ +

+ =

The coefficients ofv

1,v

2 are zero as

1 1 2 2

( ) ( )y v y x v y x = +


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3-Sep-12 9

y1,y2 are solutions of the associatedhomogeneous l.d.e.

0 1 2( ) ( ) ( ) 0a x y a x y a x y + + =

Thus we get

1 1 2 2

0

( )( ) ( ) ....(2)( )

h xv y x v y xa x

+ =

Solving (1), (2) we get 1 2,v v

Integrating, we get 1 2,v v


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3-Sep-12 10

The two equations satisfied by 1 2,v v

are1 1 2 2( ) ( ) 0 ...........(1)v y x v y x + =

1 1 2 2

0

( )( ) ( ) ....(2)

( )

h xv y x v y x

a x + =

We note that the determinant of the coefficient

matrix is

1 2

1 2

( ) ( )

( ) ( )

y x y x

y x y x=

1 2[ , ]( )W y y x 0


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3-Sep-12 11

asy1,y2 are LI solutions of the associated

homogeneous l.d.e. Using Cramers rule, we get

2

0 2

1

1 2

0 ( )

( ) / ( ) ( )

[ , ]( )

y x

h x a x y xv

W y y x

=

2

0

1 2

( )( )( )

[ , ]( )

h xy xa x

W y y x

=

1

1 0

2

1 2

( ) 0

( ) ( ) / ( )[ , ]( )

y x

y x h x a xvW y y x

=

1

0

1 2

( )( )

( )[ , ]( )

h xy x

a xW y y x

=


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3-Sep-12 12

Integrating, we get

2

01

1 2

( )( )

( )[ , ]( )

h xy x

a xv dxW y y x

=

1

02

1 2

( )( )

( )

[ , ]( )

h xy x

a xv dxW y y x=


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3-Sep-12 13

And hence a particular solution is

1 1 2 2( ) ( )py y v y x v y x= = +

( ) ( )( )

( ) ( )( )

2 10 0

1 2

( ) ( )

( ) ( )

h x h x

y x y xa x a xy x dx y x dx

W x W x

= +


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3-Sep-12 14

( ) ( )

( )

( ) ( )

( )

2 1

0 0

1 2

( ) ( )

( ) ( )x x

a a

h t h t y t y ta t a t

y x dt y x dtW t W t

= +

[ ]1 2 2 1

0

( ) ( ) ( ) ( )( )( ) ( )

x

a

y t y x y t y xh t dta t W t

= ( , ) ( )x

aK x t h t dt=

where [ ]1 2 2 10

( ) ( ) ( ) ( )( , )( ) ( )

y t y x y t y xK x ta t W t

=


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3-Sep-12 15

Example 1 Find the general solution of thed.e. 4 sec2y y x + =

Auxiliary Equation 2 4 0m + =

Roots: m = 2iHence the complementary function is

1 2cos 2 sin 2hy y c x c x= = +c1, c2 arbitrary constants


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3-Sep-12 16

Hence we take a particular solution as

1 2cos 2 sin 2y v x v x= +

where v1(x), v2(x) are functions ofx to bechosen such that the above is a solution of

the given d.e.Differentiating w.r.t.x, we get

1 2( 2sin 2 ) (2cos 2 )y v x v x = +

1 2cos 2 sin 2v x v x + +


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3-Sep-12 17

We now choose v1, v

2such that

1 2cos 2 sin 2 0 .....(1)v x v x + =

Differentiatingy

w.r.t.x, we get

1 2( 2sin 2 ) (2cos 2 )y v x v x = +

1 2( 4cos2 ) ( 4sin 2 )y v x v x = +

1 2( 2 sin 2 ) (2 cos 2 )v x v x + +


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3-Sep-12 18

Substituting fory,y,y

in the given d.e.,

we get4 sec2y y x + =

1 2( 4cos 2 ) ( 4sin 2 )v x v x +

1 2( 2sin2 ) (2cos2 )v x v x + +

1 2

4( cos2 sin 2 ) sec2v x v x x+ + =

1 2( 2sin 2 ) (2cos2 ) sec2v x v x x + =i.e.


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3-Sep-12 19

The two equations satisfied by 1 2,v v

are

1 2cos 2 sin 2 0 ..... (1)v x v x + =

1 2( 2sin 2 ) (2cos 2 ) sec2 ...(2)v x v x x + =

Using Cramers rule, we get


matrix is

cos 2 sin 2

2sin 2 2cos 2

x x

x x = 2 0


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3-Sep-12 20

1

0 sin 2sec 2 2cos 2

2

x

x xv =

1tan2

2x=

2

cos 2 0

2sin 2 sec 2

2

x

x x

v

=

1

2=


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3-Sep-12 21

Integrating, we get

1ln(sec2 )

4x

2v =

1ln(cos 2 ),

4x=

1v =

1

2x

Hence a particular solution is

1 2cos 2 sin 2y v x v x= +

1 1[ln(cos 2 )]cos 2 sin 2

4 2x x x x= +


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3-Sep-12 22

Hence the general solution is

1 2. . cos 2 sin 2i e y c x c x= + +

c1, c2 arbitrary constants

h py y y= +

1 1[ln(cos2 )]cos2 sin2

4 2

x x x x+ +


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3-Sep-12 23

Alternatively, here we can directly find v1 andv2 by

20

1

1 2

( )

( )( )

[ , ]( )

h x

y xa xv dx

W y y x

=

10

2

1 2

( )( )

( )

[ , ]( )

h xy x

a xv dx

W y y x=


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3-Sep-12 24

Example 2 Find the general solution of thed.e. 2 2( 1) (2 ) (2 ) ( 1)x x y x y x y x + + + = +

We first find the complementary function .

1

xy y e= =

We assume a second LI solution as

As the sum of the coefficients (of the LHS)

2( 1) (2 ) (2 ) 0x x x x= + + + =

is one solution of the C.F.

2 1y y vy= = where


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3-Sep-12 25

2

1

1 P dxv e dxy

= =2(2 )

( 1)

21

xdxx x

xe dx

e

+

Now2 2

( 1)

x

x x

=

+

21

( 1)

x

x x

+ =

+

2 11

1x +

+

Hence 2(2 )( 1)

xdx

x xe

+=

2 1(1 )1

dxx xe

++ 2

1 xx ex

+=


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3-Sep-12 26

2(2 )

( 1)21

xdx

x xxv e dx

e

+

= Thus

2 21 1x

xx e dx

e x+= 21 1( )

xe dxx x

= + 1xe

x=

Hence 2 1y vy= = 1x

And hence 21

y x= is a second LI solution

of the associated homogeneous d.e.


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3-Sep-12 27

Hence the C.F. is1 2

1xy c e c

x= +

(c1, c2 arbitrary constants)

So let a particular solution be 1 21x

y v e v x= +

Differentiating w.r.t.x, we get

1 2 1 221 1( )x xy v e v v e vx x

= + + +

We now choose v1, v2 such that

1 2

10 ....(1)xv e v

x + = Hence 1 2 2

1( )xy v e v

x = +


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3-Sep-12 28

Differentiatingy w.r.t. x, we get

1 2 1 23 2

2 1( ) ( )x xy v e v v e vx x

= + + +

Substituting fory,y,y in the given d.e.,

1 2 1 23 2

2 1( 1)[ ( ) ( )]x xx x v e v v e vx x + + + +

2

1 2

1(2 )[ ] ( 1)xx v e v x

x + + = +

2

1 2 2

1(2 )[ ( )]xx v e v

x+ +

1 2 2

1( )xy v e v

x = +

we get


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3-Sep-12 29

1 2 2

1 ( 1)

( ) ....(2)

x x

v e v x x

+ + =

The two equations satisfied by 1 2,v v are

1 21 0 ....(1)xv e vx

+ =

1 2 21 ( 1)( ) ....(2)x xv e v

x x+ + =


matrix is2

1 1xex x

+

2

1x xex

+ =


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3-Sep-12 30

Using Cramers rule, we get

2

1

2

10

( 1) 1

1( )x

x

x

x xv

xex

+

=+

x

e

=

2

2

0

1

1( )

x

x

x

e

xex

vx

e x

+

=+

x=

Integrating, we get 1v = ,x

e

2v =2

2

x


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3-Sep-12 31


1 21xy v e vx

= + 12x=

2

1 2, 2

x x

v e v

= =

Hence the general solution is h py y y= +

i.e.1 2

1xy c e c

x= + 1

2

x

c1, c2 arbitrary constants


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3-Sep-12 32

Example 3 Find the general solution of thed.e.

2 ln .xy y y e x + + =

Auxiliary Equation 2 2 1 0m m+ + =

Roots: m = 1, 1

Hence the complementary function is

1 2( )x xhy y c e c xe = = +c1, c2 arbitrary constants


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3-Sep-12 33

Hence we take a particular solution as

1 2

x xy v e v xe

= +

where v1(x), v2(x) are functions ofx to bechosen such that the above is a solution of the

given d.e.

Thus here 1 2,x xy e y xe = =

Wronskian = W =x x

x x xe xee e xe

=

2xe


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3-Sep-12 34

Noting that 0( ) 1, ( ) lnx

a x h x e x

= = we get

2

01

1 2

( )( )

( )[ , ]( )

h xy x

a xv dxW y y x

=

1

02

1 2

( )( )

( )[ , ]( )

h xy x

a xv dxW y y x

=

2ln

x x

xe x xe dxe

=

lnx x dx=

2 2ln

2 4

x xx= +

2lnx x

x

e x e dxe

=

ln x dx= lnx x x=


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3-Sep-12 35


1 2

x x

py y v e v xe

= = +

2 2

( ln )2 4

xx xx e

= + + ( ln )xx x x xe

223( ln )

2 4xx x x e=

And the general solution ish p

y y y= +

i.e. 1 2x x

y c e c xe

= + +2

23( ln )2 4

xxx x e


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3-Sep-12 36

Example 4 Find the general solution of thed.e. 2 2 2 .xx y x y y xe + =

Solution Consider the associatedhomogeneous equation

zx e=

Note that2

2,

dy d y dyxy xy

dz dz dz = =

2

2 2 0x y x y y + =

We put

(**)


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3-Sep-12 37

Auxiliary equation2

3 2 0m m + =Roots m = 1, 2

Solution of Eqn (**) is

2

1 2

z z

y c e c e= +

21 2y c x c x= +c1, c2 arbitrary constants

i.e. The complementary function of the

given d.e. is

Hence the equation (**) becomes2( 3 2) 0y + = ( )d

dz =


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3-Sep-12 38

So we take a particular solution as21 2y v x v x= +

Thus here 21 2,y x y x= =

Wronskian = W =2

1 2

x x

x=

2x


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3-Sep-12 39

Noting that2

0( ) , ( )x

a x x h x xe

= =

We get

20

1

1 2

( )

( )( )

[ , ]( )

h x

y xa xv dx

W y y x

= 1

02

1 2

( )( )

( )[ , ]( )

h xy x

a xv dxW y y x

=

xe

dx

x

= 2xe

dxx

=

x xe e

dxx x

=


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3-Sep-12 40


21 2py y v x v x= = +

2x x

xe e

x dx xe x dxx x

=

And the general solution is h py y y= +

i.e.2

1 2y c x c x= +

2

( )

xx e

xe x x dxx

= +

2( )x

x exe x x dx

x

+

* * *

variation of parameters

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