I was solving a problem stated in this question. The solution led me to the differential equation:
$$r^2R(r)'' + [r^2k^2-l(l+1)]R = 0,$$
which arises after solving $c^2\nabla^2v = v_t$ separating variables via $v=\dfrac{R(r)}{r}\Theta(\theta)T(t)$. When I put the DE in WolframAlpha I get the solution $$R(r) = c_1 \sqrt r J_{m+1/2}(\sqrt k r)+c_2 \sqrt r Y_{m+1/2}(\sqrt k r),$$
where $J_{m+1/2}$ and $Y_{m+1/2}$ are the Bessel functions of first and second kind.
I got the equation like this. We have to solve $$\frac{1}{r}\dfrac{\partial}{\partial r}(rv)+\frac{1}{r^2\sin\theta}\dfrac{\partial}{\partial\theta}\left(\sin\theta\dfrac{\partial v}{\partial\theta}\right) = \frac{1}{c^2}\dfrac{\partial v}{\partial t},$$ $$\frac{1}{R}\frac{d^2R}{dr^2} + \frac{1}{\Theta r^2\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) = \frac{1}{c^2 T}\frac{dT}{dt},$$ \begin{align} \frac{1}{R}\frac{d^2R}{dr^2} + \frac{1}{\Theta r^2\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) +\kappa^2 = 0, && \frac{1}{c^2 T}\frac{dT}{dt} +\kappa^2 = 0. \end{align}
Here we get two equations \begin{align} \frac{dT}{dt}+\kappa^2c^2T = 0. \end{align} \begin{align} \frac{r^2}{R}\frac{d^2R}{dr^2} + \frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) +\kappa^2r^2 = 0. \end{align} From the second one we separate variables: \begin{align} \frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\mu =0, && \frac{r^2}{R}\frac{d^2R}{dr^2}+\kappa^2r^2 - \mu = 0. \end{align}
To solve the angular part with Legendre polynomials we need $\mu =\ell(\ell+1)$, $\ell\in\mathbb{Z}$. And the equation I'm having trouble with is the radial one.
I would like to ask if someone could recommend a change of variable to get the solution or some deduction of the solution of the equation. I'm just starting to work with Bessel, Legendre, Mathieu functions and I don't recognise the patterns so well. I also appreciate your help.
The change $R(r)=\sqrt{r}\,\phi(k\,r)$ transforms the equation into $$ 4\,k^2\,r^2\phi''(k\,r)+4\,k\,r\,\phi'(k\,r)+(4 k^2 r^2-4 l^2+4 l-1)\,\phi(k\,r)=0. $$ Now let $k\,r=\rho$ and divide by $4$ to get Bessel's equation $$ \rho^2\phi''(\rho)+\rho\,\phi'(\rho)+\Bigl(\rho^2-\Bigl(l-\frac12\Bigr)^2\Bigr)\,\phi(\rho)=0. $$