# Laplace transform of a differential equation

Given the Laplace transform \begin{align} \mathcal{L}\{g(r)\} = f(t) = \int_{0}^{\infty} e^{-tr} g(r) \ dr \end{align} can it be shown that the transform of the differential equation \begin{align} \left[ \frac{d^{2}}{dr^{2}} - (2 \mu_{3} -1) \frac{1}{r} \frac{d}{dr} - \mu_{1}^{2} r - \mu_{2}^{2} \right] g(r) = 0 \end{align} is \begin{align} (t^{2} - \mu_{1}^{2}) \frac{df}{dt} + [ (2\mu_{3}+1) t + \mu_{2}^{2} ] f = 0 \end{align} with solution \begin{align} f(t) = A (t+\mu_{1})^{-2 \mu_{3}-1} \ \left( \frac{t+\mu_{1}}{t - \mu_{1}} \right)^{p} \end{align} where $p$ is to be determined?

Notes: This question is proposed based on the paper by A. Arda et al for which this is from equations (12)-(17). The answer they obtain once the inverse transform is taken is indeed correct, but the method presented in the above problem seems elusive, at present time.

• Are you sure that is the transform? I'm getting something else, but maybe I went wrong somewhere. Please see my answer. Commented Jul 2, 2014 at 6:32
• @M.Vinay I have added reference to the paper of which the problem stems. The answer you have presented below is along the second order differential equation I obtained when I worked the problem as well. Commented Jul 2, 2014 at 22:45

$\mathcal L[g''(r)] = t^2f(t) - t\,g(0) - g'(0) \Rightarrow\\ \mathcal L[r\,g''(r)] = \dfrac{d}{dt}[t^2f(t) - t\,g(0) - g'(0)] = t^2f'(t) + 2t\,f(t) - g(0)$

$\mathcal L[g'(r)] = tf(t) - g(0)$

$\mathcal L[r\,g(r)] = -f'(t), \mathcal L[r^2 g(r)] = f''(t)$

Rewriting the given equation as

$r\,g''(r) - (2\mu_3 - 1) g'(r) - \mu_1^2r^2 g(r) - \mu_2^2 r\,g(r) = 0$

and using the above:

$t^2 f'(t) + 2t\, f(t) - g(0) - (2\mu_3 - 1)(t\,f(t) - g(0)) - \mu_1^2 f''(t) + \mu_2^2 f'(t) \Leftrightarrow\\$

$\boxed{\mu_1^2 f''(t) - (t^2 + \mu_2^2)f'(t) + (2\mu_3 - 1)t\,f(t) = 2\mu_3g(0)}$