Lecture Notes in Differential Equations - Bruce E. Shapiro

145
Therefore since z = y ′ ,
Integrating gives
dy
dt = 1 6 t − 1 36 + Ce−6t (17.19)
y = 1
12 t2 − 1 36 t − 1 6 Ce−6t + C ′ (17.20)
It is customary to combine the (−1/6)C into a single unknown constant,
which we again name C,
y = 1 12 t2 − 1 36 t + Ce−6t + C ′ (17.21)
Comparing this with (17.12) and (17.13) we see that a particular solution
is
y P = 1 12 t2 − 1
(17.22)
36
This method also works when b(t) has t dependence.
Example 17.3. Solve
Substituting z = y ′ gives
⎫
y ′′ + 2ty ′ = t⎪⎬
y(0) = 1
⎪⎭
y ′ (0) = 0
An integrating factor is exp (∫ 2tdt ) = e t2 . Hence
(17.23)
z ′ + 2tz = t (17.24)
d
(
ze t2) = te t2 (17.25)
dt
∫
ze t2 = te t2 dt + C = 1 2 et2 + C (17.26)
z = 1 2 + Ce−t2 (17.27)
From the initial condition y ′ (0) = z(0) = 0, C = −1/2, hence
dy
dt = 1 2 − 1 2 e−t2 (17.28)
Changing the variable from t to s and integrating from t = 0 to t gives
∫ t
0
∫
dy t
ds ds = 1
2 ds − 1 2
0
∫ t
0
e −s2 ds (17.29)