Lecture Notes in Differential Equations - Bruce E. Shapiro

195
where
(∫
ν(t) = exp
)
−r 2 dt = e −r2t (22.79)
Using (22.77) in (22.78),
y(t) = 1 ∫
p(t)ν(t)dt + C 2
ν(t)
ν(t)
= 1 ∫ [ (∫ 1
ν(t) µ(t)
= 1
ν(t)
= 1
ν(t)
∫ [ (∫ ν(t)
q(t)µ(t)dt
µ(t)
∫ (∫ )
ν(t)
q(t)µ(t)dt
µ(t)
)]
q(t)µ(t)dt + C 1 ν(t)dt + C 2
ν(t)
)
+ C 1ν(t)
µ(t)
dt + 1
ν(t)
]
dt + C 2
ν(t)
∫
C1 ν(t)
µ(t) dt + C 2
ν(t)
(22.80)
(22.81)
(22.82)
(22.83)
Substituting µ = e −r1t and ν = e −r2t ,
∫ (∫
) ∫
y(t) = e r2t e (r1−r2)t q(t)e −r1t dt dt + C 1 e r2t e (r1−r2)t dt + C 2 e r2t
If r 1 ≠ r 2 , then
and thus (when r 1 ≠ r 2 ),
y(t) = e r2t ∫
e (r1−r2)t (∫
∫
e (r1−r2)t dt =
(22.84)
1
r 1 − r 2
e (r1−r2)t (22.85)
)
q(t)e −r1t dt dt + C 1 e r1t + C 2 e r2t (22.86)
Note that the C 1 in (22.86) is equivalent to the C 1 in (22.88) divided by
(r 1 − r 2 ); since C 1 is arbitrary constant, the r 1 − r 2 has been absorbed into
it.
If r 1 = r 2 = r in (22.88) (this occurs when B 2 = 4C and hence r = −B/2 =
−b/2a), then
∫
e (r1−r2)t dt = t (22.87)
hence when r 1 = r 2 = r,
y(t) = e rt ∫ (∫
)
q(t)e −rt dt dt + (C 1 t + C 2 )e rt (22.88)
Thus the homogeneous solution is
{
C 1 e r1t + C 2 e r2t if r 1 ≠ r 2
y H =
(C 1 t + C 2 )e rt if r 1 = r 2 = r
(22.89)