Ch03 7
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The variation of parameters method is used to find a particular solution to a nonhomogeneous differential equation, given solutions to the associated homogeneous equation. This method involves finding expressions for u1 and u2 by solving two equations, then the particular solution is a linear combination of the fundamental solutions involving u1 and u2. The general solution is the particular solution plus the general solution to the homogeneous equation.
![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
t
tutu
2sin
2cos
)()( 12
′−=′
( ) ( )
( ) ( )[ ]
ttu
t
tt
tttu
ttttuttu
tt
t
t
tuttu
cos3)(
sin
cossin2
32cos2sin)(2
2sincsc32cos)(22sin)(2
csc32cos
2sin
2cos
)(22sin)(2
1
22
1
2
1
2
1
11
−=′
=+′−
=′−′−
=
′−+′−
02sin)(2cos)(
csc32cos)(22sin)(2
21
21
=′+′
=′+′−
ttuttu
tttuttu](https://image.slidesharecdn.com/ch037-150731094629-lva1-app6892/85/Ch03-7-5-320.jpg)
![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
2sin2cos2sincotcscln
2
3
sin3
)(2sincotcscln
2
3
1cos2sincossin23
)(2sincotcscln
2
3
2cossin2sincos3
)(2sincos32sincotcscln
2
3
2cossin3
)(2sin)(2cos)()(
21
22
21
++−+=
+−+−−=
+−+−=
++−+−=
++=
tctctytyy C 2sin2cos)(,csc34 21 +==+′′](https://image.slidesharecdn.com/ch037-150731094629-lva1-app6892/85/Ch03-7-7-320.jpg)
Ch03 7
- 1. Ch 3.7: 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
- 2. 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 sint or cost, 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 ′+′++−=′
- 3. Example: Derivatives, 2nd Equation (2 of 6) From the previous slide, Note that we need two equations to solve for u1 and u2. The first equation is the differential equation. To get a second equation, we will require Then Next, ttuttuttuttuty 2sin)(2cos)(2cos)(22sin)(2)( 2121 ′+′++−=′ 02sin)(2cos)( 21 =′+′ ttuttu ttuttuty 2cos)(22sin)(2)( 21 +−=′ ttuttuttuttuty 2sin)(42cos)(22cos)(42sin)(2)( 2211 −′+−′−=′′
- 4. 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 =++ −′+−′− 02sin)(2cos)( csc32cos)(22sin)(2 21 21 =′+′ =′+′− ttuttu tttuttu tttuttu csc32cos)(22sin)(2 21 =′+′− tyy csc34 =+′′
- 5. 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 t tutu 2sin 2cos )()( 12 ′−=′ ( ) ( ) ( ) ( )[ ] ttu t tt tttu ttttuttu tt t t tuttu cos3)( sin cossin2 32cos2sin)(2 2sincsc32cos)(22sin)(2 csc32cos 2sin 2cos )(22sin)(2 1 22 1 2 1 2 1 11 −=′ =+′− =′−′− = ′−+′− 02sin)(2cos)( csc32cos)(22sin)(2 21 21 =′+′ =′+′− ttuttu tttuttu
- 6. Example : Solve for u1 and u2 (5 of 6) From the previous slide, Then Thus tt t t t t t tt t t t t ttu sin3csc 2 3 sin2 sin2 sin2 1 3 sin2 sin21 3 cossin2 sin21 cos3 2sin 2cos cos3)( 2 22 2 −= −= − = − = =′ 222 111 cos3cotcscln 2 3 sin3csc 2 3 )()( sin3cos3)()( ctttdtttdttutu cttdtdttutu ++−= −=′= +−=−=′= ∫∫ ∫∫ t t tututtu 2sin 2cos )()(,cos3)( 121 ′−=′−=′
- 7. 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 2sin2cos2sincotcscln 2 3 sin3 )(2sincotcscln 2 3 1cos2sincossin23 )(2sincotcscln 2 3 2cossin2sincos3 )(2sincos32sincotcscln 2 3 2cossin3 )(2sin)(2cos)()( 21 22 21 ++−+= +−+−−= +−+−= ++−+−= ++= tctctytyy C 2sin2cos)(,csc34 21 +==+′′
- 8. 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 Doing so, and using the Wronskian, we obtain Thus )()()()()( 0)()()()( 2211 2211 tgtytutytu tytutytu =′′+′′ =′+′ )()()()()( )()()( 2211 tytutytuty tgytqytpy += =+′+′′ ( ) ( ) )(, )()( )(, )(, )()( )( 21 1 2 21 2 1 tyyW tgty tu tyyW tgty tu =′−=′ ( ) ( )∫∫ +=+−= 2 21 1 21 21 2 1 )(, )()( )(, )(, )()( )( cdt tyyW tgty tucdt tyyW tgty tu
- 9. 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 ( ) ( )∫∫ +−= dt tyyW tgty tydt tyyW tgty tytY )(, )()( )( )(, )()( )()( 21 1 2 21 2 1 )()()()( 2211 tYtyctycty ++= )2(0)()( )1()()()( =+′+′′ =+′+′′ ytqytpy tgytqytpy
- 10. 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 ( ) ( )∫∫ +−= dt tyyW tgty tydt tyyW tgty tytY )(, )()( )( )(, )()( )()( 21 1 2 21 2 1 )()()()( 2211 tYtyctycty ++= )2(0)()( )1()()()( =+′+′′ =+′+′′ ytqytpy tgytqytpy