Lecture Notes in Differential Equations - Bruce E. Shapiro

35
Substituting y = t = 0 in this immediately yields C = 0. This problem
is a direct consequence of the fact that we divided our equation through
by t 4 previously to get an express solution for y(t) (see the transition from
equation 4.72 to equation 4.73): this division is only allowed when t ≠ 0.
Example 4.7. Solve the initial value problem
dy
dt − 2ty = √ 2 ⎫
⎬
π
⎭
y(0) = 0
(4.83)
Since p(t) = −2t, an integrating factor is
(∫ )
µ = exp −2tdt = e −t2 (4.84)
Following our usual procedure we get
d
(
ye −t2) = √ 2 e −t2 (4.85)
dt
π
If we try to solve the indefinite integral we end up with
ye −t2 = √ 2 ∫
e −t2 dt + C (4.86)
π
Unfortunately, there is no exact solution for the indefinite integral on the
right. Instead we introduce a new concept, of finding a definite integral.
We will use as our lower limit of integration the initial conditions, which
means t = 0; and as our upper limit of integration, some unknown variable
u. Then ∫ u
Then we have
0
d
(
ye −t2) dt =
dt
∫ u
(
ye −t2) (
− ye −t2) = 2 ∫ u
√
t=0 t=u π
0
2
√ π
e −t2 dt (4.87)
Using the initial condition y(0) = 0, the left hand side becomes
0
e −t2 dt (4.88)
y(u)e −(u)2 − y(0)e −(0)2 = ye −u2 (4.89)
hence
ye −u2 = √ 2 ∫ u
e −t2 dt (4.90)
π
0