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