Second order ODE of unknown format problem I'm looking for a Real function $R(r)$
that satisfies:
$$r^2R''+R'+m^2 rR=0$$
where $m\in\Bbb R$.
It looks a bit like an Euler DE but it isn't, and a bit like a Bessel DE but isn't either.
Wolfram alpha (link to the ODE) doesn't recognise it and provides no solution unfortunately.
I think I need a substitution like $r=f(u)$ to get going.
Any help is appreciated.
A little background -
The equation is the radial equation of two ODEs obtained after separation of variables, of a heat conduction problem (a very thin disk of radius $R_1$).
Boundary conditions are:
$$R(R_0)=0$$
$$R'(R_1)=0$$
I upvoted the first answer because it looked like a good idea but it turned out to be incorrect, as I showed in my comment.
In response to commenter 'tomasliam', the Sturm Liouville form of the DE is:
$$\frac{\text{d}}{\text{d}r}\left[e^{-1/r}R'(r)\right]+\frac{e^{-1/r}m^2R(r)}{r}=0$$

On request of @themaker:
A very thin disc of radius $R_1$ is at temperature $T_i$. It is insulated on both sides, as well as the outer edge.
At $t=0$ the area $[0,R_0]$ is suddenly heated to $T_0$.
What is the temperature evolution $T(t,r)$ of the disc (on $[R_0,R_1]$)?
Fourier heat equation for the disc, taking symmetry into account:
$$T_t=\frac{\alpha}r\frac{\partial}{\partial r}\Big(r\frac{\partial T}{\partial r}\Big)$$
$$\frac{T_t}{\alpha}=\frac1r(T_{r}+rT_{rr})$$
For homogeneity, we make a substitution:
$$u(t,r)=T(t,r)-T_0$$
$$\frac{u_t}{\alpha}=\frac1r(u_{r}+ru_{rr})$$
Initial:
$$u(0,r)=T_i-T_0$$
Boundaries:
$$u(t,R_0)=0$$
$$u_r(t,R_1)=0$$
Ansatz:
$$u(t,r)=\Theta (t)R(r)$$
Substitute, then divide by $u$:
$$\frac{\Theta'}{\alpha \Theta}=\frac{1}{r}\frac{R'}R+\frac{R''}R=-m^2$$
$$\frac{1}{r}\frac{R'}R+\frac{R''}R=-m^2$$
$$rR''+R'+m^2 rR=0$$
So it looks an error was made in setting up the original ODE! Mea culpa.
The solution of the last equation is:
$$R(r)=c_1J_0(mr)+c_2Y_0(mr)$$
 A: HINT.
My attempts to get exact solution have not bring useful results.
At the same time, looks possible to get solution of the Sturm-Liouville equation in the series form.
Let 
$$E = e^{-{^1/_{\large r}}},\quad F=m^2ER',\quad v=\dfrac1{m^2},\tag1$$
then
$$rF' + ER = 0.\tag2$$
Denote
$$G_0(r) = \dfrac1rF',\quad G_{n+1} = r^2(G_n\!\!'+vF),\tag3$$
then
\begin{align}
&G_0 = rF' = - ER,\\
&G_0\!\!'= - E\left(R'+\dfrac1{r^2}R \right) =-vF-\dfrac1{r^2}ER,\quad
G_1\!\! = r^2(G_0\!\!'+vF) = -ER,\\
&G_1\!\!'= - E\left(R'+\dfrac1{r^2}R \right) =-vF-\dfrac1{r^2}ER,\quad
G_2\!\! = r^2(G_1\!\!'+vF) = -ER,\dots\\
&G_n = -ER,\quad n= 0,1,\dots.\tag4
\end{align}
Assuming
$$R(\rho_0) = 0,\quad R'(\rho_0) = q,\quad 
vF(\rho_0)=qe^{-{^{\large1\!}/{ \rho^\,_0}}},\quad R'(\rho_1)=0,\tag5$$
one can get
\begin{align}
&G_0\!\!'(\rho_0) = rF'(\rho_0) = 0,\quad F'(\rho_0) = 0,\tag{6.1}\\[4pt]
&G_1\!\!' = r^2\big((rF')'+vF\big) = r^3F''+r^2F'+vr^2F,\\[4pt]
&G_1\!\!'(\rho_0) = \rho_0^3F''(\rho_0)+\rho_0^2 qe^{-{^{\large1\!}/{ \rho^\,_0}}}= 0,\\[4pt]
&F''(\rho_0) = -\dfrac q{\rho_0}\,e^{-{^{\large1\!}/{ \rho^\,_0}}},\tag{6.2}\\[4pt]
&G_2\!\!' = r^2\big((r^3F''+r^2F'+vr^2F)'+vF\big)\\[4pt]
&= r^5F'''+4r^4F''+r^2(2r+v)F'+r^2(2r+1)vF,\\[4pt]
&G_2\!\!'(\rho_0) = \rho_0^5F'''(\rho_0)+\rho_0^2(1-2\rho_0)
qe^{-{^{\large1\!}/{ \rho^\,_0}}}= 0,\\[4pt]
&F'''(\rho_0) = \dfrac {2\rho_0-1}{\rho_0^3}\,qe^{-{^{\large1\!}/{ \rho^\,_0}}},\dots\tag{6.3}\\[4pt]
\end{align}
This recurrent process should obtain Taylor series for $F(r)$ and then for $R(r).$
Possible problem is applying of the condition to the derivative.
A: If we take $R=e^{f(r)}$ we get:
$$e^{f(r)}(r^2f’’(r)+2f’(r)+m^2r)$$
After a little algebraic manipulation and use of $e^{f(r)}$ never equaling 0:
$$r^2f’’(x)+2f’(r)=-m^2r$$
By integrating and use of integration by parts you might be able to solve it from there.
A: I have decided to contribute a numerical solution to your equation. I have the derivation of the numerical procedure and python code in jupyter. I can share both if there is interest.
I have performed finite difference discretization of the equation. I have assumed


*

*$n = 1000$ - the number of discretization points

*$r_0 = 1$ - the inner edge of the disc

*$r_n = 5$ - the outer edge of the disc


Note it is not possible to naively place the inner edge at zero because of singularity.
Using finite differences I have converted the problem into a linear discrete eigenvalue problem for a 3-diagonal matrix.

Here is a plot of sorted eigenvalues $m$ (not $m^2$) by index for progressively improving discretization. It is not unreasonable to hypothesize that the in continuum limit the eigenvalues are a linear function of their index.



These are the first three eigenvectors. As expected, the following eigenvectors continue oscillating more and more.
