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Differential Equations with Polynomial Coefficients
.

Thus,
2 1 C 2
F(s) = − + e s /2 .
s 3 s s 3
Since F(s) → 0 as s → ∞, we must have C = 0 and

2
y(t) = t − 1,

. . . . which can be verified to be the solution.
There are pitfalls, however, of which the reader should be aware. A seemingly

innocuous problem such as
′
y − 2ty = 0, y(0) = 1,
2
t
has y(t) = e as its solution, and this function, as we know, does not possess a
Laplace transform. (See what happens when you try to apply the Laplace
transform method to this problem.)
Another caveat is that if the differential equation has a regular singular point,
one of the solutions may behave like log t as t → 0+; hence its derivative has no

Laplace transform. In this case, the Laplace transform method can deliver only
.
the solution that is bounded at the origin.

Example 5.15, Solve

ty + y + 2y = 0.
′′
′
. . . . .
.
Solution. The point t = 0 is a regular singular point of the equation. Let
us determine the solution that satisfies y(0) = 1. Taking the Laplace

transform,
2
(−s F (s) − 2sF(s) + 1) + (sF(s) − 1) + 2F(s) = 0,
′
that is,
2
′
−s F (s) − sF(s) + 2F(s) = 0,
or ( )
1 2
′
F (s) + − F(s) = 0, s > 0,
s s 2
Then the integrating factor is

∫ 1 2
µ(s) = e ( − )ds = se 2/s .
s 2
s
Therefore,
( ) ′
F(s)se 2/s = 0

and
Ce −2/s
F(s) = .
. . . . s

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