I would like to know if the following differential operator on $(0,\infty)$ is well-known or derived from such one:
\begin{align}
L := \frac{1}{2}\frac{d^{2}}{dx^{2}} - a x^{b} \frac{d}{dx} \quad (a,b > 0).
\end{align}
I also would like to know if the eigenfunctions are represented by some special functions.
Thank you for any comments.



Is a differential operator $\frac{1}{2}\frac{d^{2}}{dx^{2}} - a x^{b} \frac{d}{dx}$ well-known?
*Comments*  

With Maple, solving $Lf = \lambda f$,


$\bullet\;$ In case $b=1$ we get Kummer functions $M$ and $U$.  

$$
f \left( x \right) =C\_1\,{{\rm M}\left({\frac {a+\lambda}{2\,a}
},\,{\frac{3}{2}},\,a{x}^{2}\right)}x+C\_2\,x{{\rm U}\left({
\frac {a+\lambda}{2\,a}},\,{\frac{3}{2}},\,a{x}^{2}\right)}
$$


$\bullet\;$ In case $b=2, a>0$ we get a Heun triconfluent function.  

$$
f \left( x \right) ={\rm HeunT} \left( -{\lambda\,\sqrt [3]{2}{3}^{{
\frac{2}{3}}}{a}^{-{\frac{2}{3}}}},3,0,{\frac {\sqrt [3]{2}{3}^{{\frac
{2}{3}}}x}{3}\sqrt [3]{a}} \right)
$$
[The other solution is by variation of parameters]


$\bullet\;$ In case $b=3$ we get Heun biconfluent functions.
$$
f \left( x \right) =C\_1\,{\rm HeunB} \left( -{\frac{1}{2}},0,{
\frac{3}{2}},{\lambda\,\sqrt {2}{\frac {1}{\sqrt {a}}}},{\frac {\sqrt 
{2}{x}^{2}}{2}\sqrt {a}} \right) \\
+C\_2\, x\;{\rm HeunB} \left( {
\frac{1}{2}},0,{\frac{3}{2}},{\lambda\,\sqrt {2}{\frac {1}{\sqrt {a}}}
},{\frac {\sqrt {2}{x}^{2}}{2}\sqrt {a}} \right)
$$


$\bullet\;$ In case $b=4$, Maple did not find this DE to be of Heun type.


[Closed form also found for $b=0, -1, -2$.]


I expect that, in general, the eigenfunctions are not named special functions.


