Ch 3.6: Variation of Parameters

Recall the nonhomogeneous equation

where p, q, g are continuous functions on an open interval I.The associated homogeneous equation is

In this section we will learn the variation of parameters method to solve the nonhomogeneous equation. As with the method of undetermined coefficients, this procedure relies on knowing solutions to homogeneous equation.

Variation of parameters is a general method, and requires no detailed assumptions about solution form. However, certain integrals need to be evaluated, and this can present difficulties.

)()()( tgytqytpy

0)()( ytqytpy

Example: Variation of Parameters (1 of 6)

We seek a particular solution to the equation below.

We cannot use method of undetermined coefficients since g(t) is a quotient of sin t or cos t, instead of a sum or product. Recall that the solution to the homogeneous equation is

To find a particular solution to the nonhomogeneous equation, we begin with the form

Then

or

tyy csc34

tctctyC 2sin2cos)( 21

ttuttuty 2sin)(2cos)()( 21

ttuttuttuttuty 2cos)(22sin)(2sin)(22cos)()( 2211

ttuttuttuttuty 2sin)(2cos)(2cos)(22sin)(2)( 2121

Example: Derivatives, 2nd Equation (2 of 6)

From the previous slide,

As a modification to the assumption of the solution, we further impose the following condition on u1 and u2 to make the above expression simpler and future the task easier:

Then, are reduced to the much simpler forms:

and,

ttuttuty 2cos)(22sin)(2)( 21

ttuttuttuttuty 2sin)(42cos)(22cos)(42sin)(2)( 2211

Example: Two Equations (3 of 6)

Recall that our differential equation is

Substituting y'' and y into this equation, we obtain

This equation simplifies to

Thus, to solve for u1 and u2, we have the two equations:

tttuttu

ttuttuttuttu

csc32sin)(2cos)(4

2sin)(42cos)(22cos)(42sin)(2

21

2211

tyy csc34

Example: Solve for u1' (4 of 6)

To find u1 and u2 , we need to solve the equations

From second equation,

Substituting this into the first equation,

t

ttutu

2sin

2cos)()( 12

ttu

t

tttttu

ttttuttu

ttt

ttuttu

cos3)(

sin

cossin232cos2sin)(2

2sincsc32cos)(22sin)(2

csc32cos2sin

2cos)(22sin)(2

1

221

21

21

11

02sin)(2cos)(

csc32cos)(22sin)(2

21

21

ttuttu

tttuttu

Example : Solve for u1 and u2 (5 of 6)

From the previous slide,

Then

Thus

ttt

t

t

t

t

tt

tt

t

tttu

sin3csc2

3

sin2

sin2

sin2

13

sin2

sin213

cossin2

sin21cos3

2sin

2coscos3)(

2

22

2

222

111

cos3cotcscln2

3sin3csc

2

3)()(

sin3cos3)()(

ctttdtttdttutu

cttdtdttutu

t

ttututtu

2sin

2cos)()(,cos3)( 121

Example: General Solution (6 of 6)

Recall our equation and homogeneous solution yC:

Using the expressions for u1 and u2 on the previous slide, the general solution to the differential equation is

tctctttt

tyttttttt

tyttttttt

tyttttttt

tyttuttuty

C

C

C

C

2sin2cos2sincotcscln2

3sin3

)(2sincotcscln2

31cos2sincossin23

)(2sincotcscln2

32cossin2sincos3

)(2sincos32sincotcscln2

32cossin3

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

21

22

21

tctctytyy C 2sin2cos)(,csc34 21

Summary

Suppose y1, y2 are fundamental solutions to the homogeneous equation associated with the nonhomogeneous equation above, where we note that the coefficient on y'' is 1.

To find u1 and u2, we need to solve the equations

)()()()()(

)()()(

2211 tytutytuty

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Summary

Solving the equations directly to obtain , or using the Wronskian, we may have an explicit expression:

Integrating ,we finally obtain , or have the explicit formula:

)()()()()(

)()()(

2211 tytutytuty

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)(,

)()()(,

)(,

)()()(

21

12

21

21 tyyW

tgtytu

tyyW

tgtytu

221

121

21

21 )(,

)()()(,

)(,

)()()( cdt

tyyW

tgtytucdt

tyyW

tgtytu

Theorem 3.7.1

Consider the equations

If the functions p, q and g are continuous on an open interval I, and if y1 and y2 are fundamental solutions to Eq. (2), then a particular solution of Eq. (1) is

and the general solution is

dttyyW

tgtytydt

tyyW

tgtytytY

)(,

)()()(

)(,

)()()()(

21

12

21

21

)()()()( 2211 tYtyctycty

)2(0)()(

)1()()()(

ytqytpy

tgytqytpy