A solution to the differential equation $Y'' + M(x^2 Y)' - x^2 Y = 0$ In a problem I'm working on relating to plasma instabilities, the following boundary value problem showed up
\begin{equation}
\frac{d^2Y}{dx^2} + M\frac{d}{dx}\left[x^2Y(x)\right] - x^2 Y(x) = 0\;\;\;\;;\;\;\;\; Y(0) = 1 \;\;\;\;;\;\;\;\; Y(\infty) = 0
\end{equation}
where $M$ is a real constant. The relevant quantity to the calculation is $Y'(0)$. This is easy enough to do numerically, and I can get it to first order in $M$ analytically, but I'd like it if I could find a full analytic expression for $Y'(0)$ in terms of $M$. The usual method of "tell Mathematica to do it" failed, so I'm wondering if anyone here has any ideas.
Note that a full functional form for $Y$ is not necessary, though obviously it would be helpful if it exists.
 A: According to Maple, this is essentially a triconfluent Heun differential equation.  In Maple's notation, the solution is
$$
Y(x) = { C_1}\,{\rm HeunT} \left( {\frac {{3}^{2/3}}{{M}^{8/3}}},3,2\,{
\frac {\sqrt [3]{3}}{{M}^{4/3}}},1/3\,{3}^{2/3}\sqrt [3]{M}x \right) {
{\rm e}^{-1/3\,{\frac { \left( {M}^{2}{x}^{2}+3 \right) x}{M}}}}\\
+{C_2}\,{\rm HeunT} \left( {\frac {{3}^{2/3}}{{M}^{8/3}}},-3,2\,{\frac 
{\sqrt [3]{3}}{{M}^{4/3}}},-1/3\,{3}^{2/3}\sqrt [3]{M}x \right) {
{\rm e}^{{\frac {x}{M}}}}
$$
Change variables to $y=(M/3)^{1/3}x$ to get
$$
{C_1}\,{\rm HeunT} 
\left(\left(\frac{3}{M^4}\right)^{2/3},3,2\left(\frac{3}{M^4}\right)^{1/3},y\right)
\exp\left(-y^3-(3/M^4)^{1/3}y\right)
\\
+{C_2}\,{\rm HeunT}
\left(\left(\frac{3}{M^4}\right)^{2/3},-3,2\left(\frac{3}{M^4}\right)^{1/3},-y\right) 
\exp\left((3/M^4)^{1/3}y\right)
$$
Numerical experiments suggest to me that the first term goes to $0$ as $x \to +\infty$ because of the exponential factor.  So your solution should be the one with $C_1=1,C_2=0$.
HeunT itself wildly oscillates as $x \to +\infty$.  The DE has an irregular singular point at $\infty$.  Maple does not know asymptotics for it as $x \to +\infty$.

Here are some references on Heun functions.  Maple says it follows the notation of [3] for HeunT.
[1] A. Ronveaux (Editor),
Heun's Differential Equations (Oxford University
Press, 1995)
[2] B. D. Sleeman and
V. B. Kuznetsov,
Digital Library of Mathematical Functions.  Chapter 31, Heun Functions.
http://dlmf.nist.gov/31
[3] A. Decarreau,  M.C. Dumont-Lepage,  P. Maroni, A. Robert, and A. Ronveaux, "Formes Canoniques de Equations confluentes de l'equation de Heun." Annales de la Societe Scientifique de Bruxelles, Vol. I-II. (1978): 53-78.
A: This is not a complet answer may lead to a proof for your boundary problem , We have : 
\begin{equation}
\frac{d^2Y}{dx^2} + M\frac{d}{dx}\left[x^2Y(x)\right] - x^2 Y(x) = 0\;\;\;\;;\;\;\;\; Y(0) = 1 \;\;\;\;;\;\;\;\; Y(\infty) = 0
\end{equation}
The above equation can be written as :$y''(x)+Mx²y'(x)+(2xM-x²)y=0$ and the latter give us two cases : 
for $M=\frac{x}{2}$ the equation  is second-order of linear differntial equation where the solution in this case is $\displaystyle y(x)=\frac{c_1x\Gamma(1/4,\frac{x^4}{8})}{2^{0.25}{\sqrt{x}}^{0.25}}$.
Now for $M\neq \frac{x}{2}$ , The coeffecients here are analytic functions and this follow legendre differential  equation using the power series function you will get : $y(x)=y_{0}\sum_{n=0}^{\infty}{a_{2n}{x}^{2n}}+y_1\sum_{n=0}^{\infty}{a_{2n+1}{x}^{2n+1}}$, just you can check this paper page 135.exercice 5.3 to get a complet steps for solving your ODE using power series 
