07 method of variation of parameters

8/3/2019 07 Method of Variation of Parameters

1/37

We assume a particular solution of

1 2cos sin

ax ax

Ly k e bx k e bx as 1 2cos sin

ax ax

py y A e bx A e bx

We assume a particular solution of

0 1( ... )

ax n

nLy e b b x b x

as 0 1( ... )ax n

p ny y e A A x A x

We also use multiply by the least power ofx

rule in case any term of the Given RHS is a

part of the C.F.


2/37

Example8 Find the general solution of

the d.e.

2 2 sinx

y y y e x Solution We first find the complementary

function.

Auxiliary Equation: 2 2 2 0m m Roots: m =

Hence theComplementary function is

1 2( cos sin ),

x

hy e c x c x c

1, c

2arbitrary constants

1 i


3/37

Now we find the particular solution.

As sinx

e x is a part of the C.F., we take a

particular solution as

1 2( cos sin )

x xy x A e x A e x

1 2 2 1[( ) cos ( ) sin ]x xDy x A A e x A A e x

1 2( cos sin )

x xA e x A e x

2

2 1[2 cos 2 sin ]x xD y x A e x A e x 1 2 2 1[2( ) cos 2( ) sin ]

x xA A e x A A e x

2

-2

1

Adding, we get


4/37

2( 2 2)D D y

2 12 cos 2 sin

x xA e x A e x

sin

x

e x gives

1 22 1, 2 0A A

1 2

1, 0

2A A or

Hence a particular solution is cos2

x

p

xy e x

Hence the general solution ish p

y y y

i.e. 1 2( cos sin )x

y e c x c x cos2

xxe x

c1, c2 arbitrary constants


5/37

The method of Variationof Parameters

In this lecture we discuss a general method offinding a particular solution of a non-

homogeneous l.d.e. (whether it is constant

coefficient equation or not).


6/37

Consider the second order l.d.e.

( ) ( ) ( )..(*)y P x y Q 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


7/37

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 bechosen such that (**) is a solution of (*).


8/37

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


9/37

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

1 1 2 2 1 1 2 2

1 1 2 2

1 1 2 2

( )

( ) ( )

v y v y v y v y

P x v y v y

Q x v y v y h x

The coefficients ofv1

, v2

are zero as

1 1 2 2( ) ( )y v y x v y x


10/37

y1,y2 are solutions of the associated

homogeneous l.d.e.

( ) ( ) 0y P x y Q x y

Thus we get

1 1 2 2( ) ( ) ( ) ....(2)v y x v y x h x

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

Integrating, we get 1 2,v v


11/37

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( ) ( ) ( ) ....(2)v y x v y x h 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


12/37

asy1,y2 are LI solutions of the associated

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

2

2

11 2

0 ( )

( ) ( )

[ , ]( )

y x

h x y x

v W y y x

2

1 2

( ) ( )

[ , ]( )

h x y x

W y y x

1

1

2

1 2

( ) 0

( ) ( )

[ , ]( )

y x

y x h xv

W y y x

1

1 2

( ) ( )[ , ]( )

h x y xW y y x


13/37

Integrating, we get 21

1 2

( ) ( )

[ , ]( )

h x y xv dx

W y y x

12

1 2

( ) ( )

[ , ]( )

h x y xv dx

W y y x

And hence a particular solution is

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


14/37

Example 1 Find the general solution of the d.e.

4 sec 2y y x Auxiliary Equation

24 0m

Roots: m = 2i

Hence the complementary function is

1 2cos 2 sin 2hy y c x c x



15/37

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 be

chosen such that the above is a solution of

the given d.e.

Differentiating w.r.t. x, we get

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

1 2cos 2 sin 2v x v x


16/37


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


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

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

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


17/37

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

we get4 sec 2y y x

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

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

1 2

4( cos 2 sin 2 ) sec 2v x v x x

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


18/37

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 ) (2 cos 2 ) sec 2 ...(2)v x v x x

Using Cramers rule, we get

We note that the determinant of the coefficientmatrix is

cos 2 sin 2

2sin 2 2 cos 2

x x

x x 2 0


19/37

1

0 sin 2

sec 2 2 cos 2

2

x

x xv

1tan2

2x

2

cos 2 0

2sin 2 sec 2

2

x

x xv

12

Integrating, we get1

ln(sec 2 )4

x 2v 1

ln(cos 2 ),4

x1

v 1

2x

1 1


20/37

Hence a particular solution is

1 2cos 2 sin 2y v x v x 1 1

[ln(cos 2 )]cos 2 sin 24 2

x x x x

Hence the general solution is

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


h py y y

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

4 2x x x x

1

1ln(cos 2 ),

4v x

2

1

2v x


21/37


2 2( 1) (2 ) (2 ) ( 1)x x y x y x y x We first find the complementary function .

1

x

y 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


22/37

2

1

1 P dxv e dx

y

2

(2 )

( 1)

2

1x

dxx x

x e dxe

Now

2

2( 1)

xx x

21

( 1)x

x x

2 111x x

Hence 2(2 )( 1)

xdx

x xe

2 1

(1 )1

dxx xe

2

1 xxe

x


23/37

2(2 )

( 1)

2

1x

dxx x

xv e dx

e

Thus

2 2

1 1 xx

xe dx

e x

2

1 1( )

xe dx

x x

1 x

ex

Hence2 1

y vy 1x

And hence 2 1yx

is a second LI solution

of the associated homogeneous d.e.


24/37

Hence the C.F. is 1 21x

y c e cx

(c1, c2 arbitrary constants)So let a particular solution be 1 2

1xy v e v

x

Differentiating w.r.t. x, we get

1 2 1 22

1 1( )

x xy v e v v e v

x x


1 2

10 ....(1)

xv e vx

Hence1 2 2

1( )

xy v e v

x

1


25/37


1 2 1 23 22 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 v

x x

2

1 2

1(2 )[ ] ( 1)

xx v e v x

x

2

1 2 2

1

(2 )[ ( )]x

x v e v x

1 2 2

1( )

xy v e vx

we get


26/37

1 2 2

1 ( 1)( ) ....(2)

x xv e v

x x

The two equations satisfied by 1 2,v v are

1 2

10 ....(1)

xv e v

x

1 2 2

1 ( 1)( ) ....(2)

x xv e v

x x

We note that the determinant of the coefficientmatrix is

2

1 1xe

x x

2

1x xe

x


27/37

Using Cramers rule, we get

2

1

2

1

0

( 1) 1

1( )

x

x

x

x xv xe

x

xe

2

2

0

1

1( )

x

x

x

e

xe

x

v xe

x

x

Integrating, we get 1v ,xe 2v

2

2

x


28/37


1 2

1xy v e v

x

12

x

2

1 2,

2

x xv e v

Hence the general solution ish p

y y y

i.e. 1 21x

y c e c x 1 2

x



29/37


2 ln .xy y y e x Auxiliary Equation

22 1 0m m

Roots: m = 1, 1 Hence the complementary function is

1 2( )

x x

hy y c e c x e



30/37

Hence we take a particular solution as

1 2

x x

y v e v xe

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

chosen such that the above is a solution of the

given d.e.

Thus here 1 2,x x

y e y xe

Wronskian = W = x x

x x x

e xe

e e xe

2xe


31/37

Noting that ( ) lnx

h x e x we get

21

1 2

( ) ( )

[ , ]( )

h x y xv dx

W y y x

121 2

( ) ( )

[ , ]( )

h x y xv dx

W y y x

2

lnx x

x

e x xedx

e

lnx x dx 2 2ln

2 4

x xx

2

lnx x

x

e x edx

e

ln x dx lnx x x


32/37


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 4

xx x x e

And the general solution is h py y y i.e.

1 2

x xy c e c x e

2

23( ln )

2 4

xxx x e


33/37

Example 4 Find the general solution of the d.e.2

2 2 .x

x y x y y x e

Solution Consider the associated homogeneous

equation

zx e

Note that2

2,dy d y dyxy xy

dz dz dz

2

2 2 0x y x y y

We put

(**)


34/37

Auxiliary equation2

3 2 0m m Roots m = 1, 2

Solution of Eqn (**) is2

1 2z zy c e c e

2

1 2y c x c x


i.e. The complementary function of the

given d.e. is

Hence the equation (**) becomes

2

( 3 2) 0y ( )

d

dz


35/37

So we take a particular solution as2

1 2y v x v x Thus here

2

1 2,y x y x

Wronskian = W =2

1 2

x x

x

2

x


36/37

Noting that ( )x

eh x

x

We get

2

1

1 2

( ) ( )

[ , ]( )

h x y xv dx

W y y x

1

2

1 2

( ) ( )

[ , ]( )

h x y xv dx

W y y x

xedx

x

2

xedx

x

x x

e edx

x x

i l l i i


37/37


2

1 2py y v x v x

2x x

xe ex dx xe x dx

x x

And the general solution is h py y y

i.e.

2

1 2y c x c x

2( )

xx exe x x dx

x

2( )

xx e

xe x x dxx

07 method of variation of parameters

Documents