Lecture Notes in Differential Equations - Bruce E. Shapiro

326 LESSON 31. LINEAR SYSTEMS
By the method of variation of parameters a particular solution is then
∫
y P = e At e At g(t)dt (31.189)
∫ ( 3
) ( )
= e At 4 e−3t + 1 4 e5t 3 8 e−3t − 3 8 e5t 5
1
2 e−3t − 1 2 e5t 1 4 e−3t + 3 dt (31.190)
4 e5t 6t
Using the integral formulas ∫ e at dt = 1 a eat and ∫ te at dt = ( t/a − 1/a 2) e at
we find
⎛
y P = e At ⎝
15
4(−3) e−3t + 5
4(5) e5t + 9 4
5
2(−3)
− 5 e−3t 2(5) e5t + 3 2
(
= e At −
3
2 e−3t + 17
50 e5t − 3t
−e −3t − 17
25 e5t − t 2 e−3t + 9t
Substituting equation (31.183) for e At
y P =
Hence
=
=
⎜
⎝
(
(
t
−3 − 1 9
4 e−3t − 9t
t
−3 − 1 9
)
20 e5t
10 e5t
) (
e −3t − 9 t
4 5 − ) 1
25 e
5t
) (
e −3t + 9 t
2 5 − ) 1
25 e
5t
⎞
⎠
(31.191)
( 3
) (
4 e3t + 1 4 e−5t 3 8 e3t − 3 8 e−5t −
3
1
2 e3t − 1 2 e−5t 1 4 e3t + 3 2 e−3t + 17
50 e5t − 3t
⎛ ( 4 e−5t −e −3t − 17
25 e5t − t 2 e−3t + 9t
3
4 e3t + 1 4 e−5t) ( − 3 2 e−3t + 17
50 e5t − 3t
4 e−3t − 9t
+ ( 3
8 e3t − 3 8 e−5t) ( ( −e −3t − 17
25 e5t − t 2 e−3t + 9t
1
2 e3t − 1 2 e−5t) ( − 3 2 e−3t + 17
50 e5t − 3t
4 e−3t − 9t
(
−
29
25 − 6 5 t
)
− 42
25 + 2 5 t
+ ( 1
4 e3t + 3 4 e−5t) ( −e −3t − 17
25 e5t − t 2 e−3t + 9t
)
4 e−3t − 9t
20 e5t
10 e5t
20 e5t)
⎞
10 e5t)
20 e5t)
⎟
⎠
10 e5t)
(31.192)
y = y H + y P (31.193)
( ) ( ) (
= c 1 e 3t 3
+ c
2 2 e −5t 1 −
29
+ 25 − 6 5 t )
−2 − 42
25 + 2 5 t (31.194)
In terms of the variables x and y in the original problem,
x = 3c 1 e 3t + c 2 e −5t − 29
25 − 6 5 t (31.195)
y = 2c 1 e 3t − 2c 2 e −5t − 42
25 + 2 5 t (31.196)