Nearly Bessel's Equation In my research I have come across a linear, second-order ordinary differential equation that looks much like Bessel's equation except for one small difference:
\begin{equation}
z^2\frac{\partial^2}{\partial z^2}R_m(z) + z\frac{\partial}{\partial z}R_m(z) + \left(z^2+Wz-m^2\right)R_m(z) = 0,
\end{equation}
where $m$ and $W$ are constants.  If $W\rightarrow0$, then Bessel's equation is recovered.  I believe this equation may be solved as a series, but its similarity to Bessel's equation makes me wonder if there is another way (maybe through some clever substitution) that makes use of well known functions.  I don't especially want to re-invent the wheel and I have tried to look up this equation, but I have not been able to find anything.  Does anyone know if this equation has previously established solutions?
 A: Just an idea I had. Let's assume $W \ll z$ (which should certainly hold for large enough $z$), then we can use perturbation methods for the equation:
$$z^2\frac{\partial^2}{\partial z^2}f + z\frac{\partial}{\partial z}f + \left(z^2-m^2\right)f+Wzf = 0$$
Assume:
$$f=f_0+W f_1+W^2f_2+\dots$$
Then, by keeping only a certain order of $W$ terms, we have a sequence of equations:
$$z^2\frac{\partial^2}{\partial z^2}f_0 + z\frac{\partial}{\partial z}f_0 + \left(z^2-m^2\right)f_0 = 0$$
This one gives Bessel functions.
$$z^2\frac{\partial^2}{\partial z^2}f_1 + z\frac{\partial}{\partial z}f_1 + \left(z^2-m^2\right)f_1+z f_0 = 0$$
This is inhomogeneous Bessel equation, which also can be solved exactly.
$$z^2\frac{\partial^2}{\partial z^2}f_2 + z\frac{\partial}{\partial z}f_2 + \left(z^2-m^2\right)f_2+z f_1 = 0$$
And so on. In case the Mathematica solution which  Dr. Wolfgang Hintze provided in the comments is not convenient to use, it's always possible to use this perturbation method for large $z$ and the (Frobenius) series method for small $z$.
A: A method to recognize the equation is to make the first order differential term vanish by changing $R_m(z)=z^{-1/2}f(z)$. We obtain
\begin{equation}
z^2f''(z)+ \left(z^2+Wz-m^2+\frac{1}{4}\right)f(z) = 0
\end{equation} 
which is very similar to the Whittaker's equation. To identify both we further change $z=\tfrac{u}{2i}$ and $f(z)=g(u)$:
\begin{equation}
g''(u)+\left(-\frac{1}{4}+\frac{W}{2i}\frac{1}{u}+\frac{\tfrac{1}{4}-m^2}{u^2}\right)g(u) = 0
\end{equation} 
The solutions of the original equation are thus given by
\begin{equation}
R_m(z)=z^{-1/2}\left[AW_{W/2i,m}(2iz)+BM_{W/2i,m}(2iz)  \right]
\end{equation} 
where $A,B$ are arbitrary constants and $W_{\kappa,\mu}$ and $M_{\kappa,\mu}$ are Whittaker functions which are confluent hypergeometric functions. Many of their properties are given in DLMF. This solution is consistent with the Mathematica expression given in comments by @Dr. Wolfgang Hintze 
