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Application of Laplace Transform
.

x
n
Taking the series e = ∑ ∞ (x /n!) with x = −2/s implies
n=0
∞
n n
∑ (−1) 2
F(s) = C .
n!s n+1
n=0
−1
We can take L term by term so that
∞ n n n
∑ (−1) 2 t
y(t) = C .
(n!) 2
n=0

The condition y(0) = 1 gives C = 1.
√
Note that y(t) = J (2 at) with a = 2, from the table of Laplace
0
transforms, where J is the well-known Bessel function. There is another
0
solution to this differential equation which is unbounded at the origin and

. . . . cannot be determined by the preceding method.

Integral Equations

Equations of the form

∫
t
f(t) = g(t) + κ(t, τ)f(τ)dτ
0

and
∫ t
g(t) = κ(t, τ)f(τ)dτ
0
are known as integral equations, where f(t) is the unknown function. When the
kernel κ(t, τ) is of the particular form

k(t, τ) = κ(t − τ),

the integrals represent convolutions. In this case, the Laplace transform lends
itself to their solution.
Considering the first type, if g and κ are known, then formally

L(f) = L(g) + L(f)L(κ)

by the convolution theorem. Then

L(g)
L(f) = ,
1 − L(κ)

and from this expression f(t) often can be found since the right-hand side is just

a function of the variable s.

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