Lecture Notes in Differential Equations - Bruce E. Shapiro

327
Non-constant Coefficients
We can solve the general linear system
y ′ = A(t)y(t) + g(t) (31.197)
where the restriction that A be constant has been removed, in the same
manner that we solved it in the scalar case. We expect
[ ∫ ]
M(t) = exp − A(t)dt
(31.198)
to be an integrating factor of (31.197); in fact, since
M ′ (t) = −M(t)A(t) (31.199)
we can determine
d
dt (M(t)y) = M(t)y′ + M ′ (t)y (31.200)
= M(t)y ′ − M(t)A(t)y (31.201)
= M(t)(y ′ − A(t)y(t)) (31.202)
= M(t)g(t) (31.203)
Hence
∫
M(t)y =
M(t)g(t)dt + C (31.204)
and therefore the solution of (31.197) is
[∫
]
y = M −1 (t) M(t)g(t)dt + C
for any arbitrary constant vector C.
Example 31.6. Solve y ′ = A(t)y + g(t), y(0) = y 0 , where
( ) ( ) ( )
0 −t
t 4
A =
, g(t) = , y
−t 0
3t 0 =
2
(31.205)
(31.206)
To find the integrating factor M = exp ∫ −A(t)dt, we first calculate
∫
P(t) = − A(t)dt = 1 ( ) 0 t
2
2 t 2 (31.207)
0
The eigenvalues of P are
λ 1 = − t2 2 , λ 2 = t2 2
(31.208)