Greg Gamble's differential equation [2] is an example of a type that may usefully be looked at by means of the Laplace transform. (For another example see [1] which discusses in this way the equation (t-1)X^{\prime\prime}(t) -tX^\prime(t) + X(t) = 0.) In the case of Gamble's equation, this is hardly the most efficient method; it is in fact most inefficient. But it does lead to some interesting results.
To proceed, I will generalise the equation slightly, somewhat alter
the notation (to use standard forms) and discuss only the
homogeneous case (as no real generality is lost thereby). Thus we
consider the equation
(t-\alpha)X^\prime(t) = kX(t) \tag1
where, for convenience, we suppose that \alpha > 0. The solution is
X(t) = a(\alpha - t)^kU(\alpha - t) + b(t - \alpha)^kU(t - \alpha) \tag2
where U(t) is the Heaviside unit step function and a,b are arbitrary.
If we write \Cal{L}\{X(t)\} = x(s), then from equation (1),
x(s) =
\alpha X(0)s^{-(k+1)}\exp(-\alpha s) \int_{0}^{s} u^k\exp(\alpha u) du
+ As^{-(k+1)}\exp{(-\alpha s)} \tag3
where A is a constant of integration. Similarly from (2),
x(s) = a\int_{0}^{\alpha}(\alpha-t)^k \exp(-st) dt
+ b\exp(-\alpha s)\Gamma(1+k)s^{-(1+k)}. \tag4
The second terms of (3) and (4) are readily seen to be equivalent
with a = b\Gamma(1+k) and so too are the first terms if we note that
X(0) = a\alpha^k. This may be verified directly via the integral
identity
\alpha^{k+1}\int_{0}^{s} u^k \exp(\alpha u) du
= s^{k+1} \int_0^\alpha \nu^k \exp(s\nu) d\nu \tag5
or else via tables (e.g. [3, II.10.24]).
\item{[2]} {Gamble, G. `A first year differential equation'. Austral. Math. Soc. Gazette 22 (1995) 168-172.}
\item{[3]} {Oberhettinger, F. and Badii, L.
Tables of Laplace Tranforms.
New York: Springer Verlag, 1973.}
Michael A. B. Deakin
Department of Mathematics
Monash University