Eigenfunctions of an operator using Laguerre Polynomials I am trying to find the eigenfunctions of the following operator:
$$\mathcal{L}f=(-\gamma x+\frac{\mu}{x})f_x+\mu f_{xx}$$
I know that I must somehow use Laguerre polynomials, the solutions to the following ODE. $L_n(x)$ solves
$$xy''+(1-x)y'+ny=0$$. 
Here is the current ODE I have which should be solved
$$\mu {\phi}_{xx}+(-\gamma x + \frac{\mu}{x})\phi_x -\lambda \phi =0$$
That operator is the generator of the following stochastic differential equation
$$dA_t=(-\gamma x + \frac{\mu}{x})dt+\sqrt{2\mu}dW_t$$
where $W_t$ is Brownian motion. I am about 90% sure this is true, but this $may$ be something I'm doing wrong.
In an example in the notes for the section in which I find this problem, the Professor finds the eigenfunctions of 
$$\mathcal{L}f=-\alpha x f_x + \frac{1}{2}\sigma^2f_{xx}$$
by solving the ODE
$$\phi '' - \frac{2\alpha x}{\sigma^2} \phi ' + \frac{2\lambda }{\sigma^2}\phi=0$$
via substitution with some $y$ to get 
$$\phi '' - y\phi ' +\frac{\lambda}{\alpha}\phi =0 $$
which is then solved with Hermite polynomials. The reason I include this is that I believe some clever substitution may be necessary in my problem as well, but I cannot figure it out! I greatly appreciate any help. 
EDIT: Based on the question, I am almost certain that the eigenvalues are $\lambda = -2\gamma n$ for positive integers $n$. Maybe this can help. 
EDIT2: For clarification, the generator of a process $dX_t=b(X_t,t)dt+\sigma(X_t,t)dW_t$ is found thus:
$$\mathcal{L}f=b(x,t)\cdot \nabla f(x)+\frac{1}{2}a(x,t):\nabla^2 f(x)$$ where $a(x,t)=\sigma \sigma^T$. This is how I found the generator, which I'm 90% sure is right. 
 A: Let $x=t^a$ then
Where
$$
\dfrac{dx}{dt} = at^{a-1}
$$
$$
\dfrac{d}{dx} = \dfrac{dt}{dx}\dfrac{d}{dt} = \frac{1}{a}t^{-a+1}\dfrac{d}{dt}\\
\dfrac{d^2}{dx^2} =    \frac{1}{a}t^{-a+1}\dfrac{d}{dt}\frac{1}{a}t^{a-1}\dfrac{d}{dt}= \frac{1}{a^2}t^{2(1-a)}\dfrac{d^2}{dt^2} + \frac{1-a}{a^2}t^{1-2a}\dfrac{d}{dt}
$$
Thus we get
$$
\frac{\mu}{a^2}t^{2(1-a)}\dfrac{d^2y}{dt^2} + \frac{\mu(1-a)}{a^2}t^{1-2a}\dfrac{dy}{dt} -\gamma t^a\frac{1}{a}t^{-a+1}\dfrac{dy}{dt} +\frac{\mu}{t^a}\frac{1}{a}t^{-a+1}\dfrac{dy}{dt}-\lambda y = 0
$$
This leads to
$$
\frac{\mu}{a^2}t^{2(1-a)}\dfrac{d^2y}{dt^2} + \frac{\mu(1-a)}{a^2}t^{1-2a}\dfrac{dy}{dt} -\gamma t\frac{1}{a}\dfrac{dy}{dt} +\mu \frac{1}{a}t^{-2a+1}\dfrac{dy}{dt}-\lambda y = 0
$$
Let $a =1/2$
Then we get
$$
\frac{\mu}{a^2}t\dfrac{d^2y}{dt^2} + \frac{\mu(1-a)}{a^2}\dfrac{dy}{dt} -\gamma t\frac{1}{a}\dfrac{dy}{dt} +\mu \frac{1}{a}\dfrac{dy}{dt}-\lambda y = 0
$$
Which sort of looks like the Lagurre poly equation, up to normalisation. (But check the calculations). 
Subbing in $a$ we find
$$
t\ddot{y} +\left(1-\frac{\gamma }{2\mu}t\right) \dot{y} -\frac{\lambda}{4\mu}y =0.
$$
Then using the transformation $t\rightarrow \beta s$
We find
$$
s\dfrac{d^2y}{ds^2} +\left(1- s\right)\dfrac{dy}{ds}  - \frac{\lambda}{2\gamma}y = 0
$$
Where $\beta = \frac{2\mu}{\gamma}$ the final statement is fixing 
$$
n = -\frac{\lambda}{2\gamma} \implies \lambda = -2\gamma n.
$$
