A linear but intractable PDE I have a PDE of the following form, from a physics problem:
$$
y \left(\alpha \frac{\partial }{\partial y}+x \frac{\partial^2 }{\partial x \partial y} \right)f(x,y) = \left( z_1 + z_2 x^\alpha y^{-2} \right) f(x,y)
$$
Function $f(x,y)$ is a real-space real-valued function and 
$z_{1,2},\alpha$ are real numbers, generally irrational.   The latter, specifically the $z_2$ coefficient term, seems to make all of the textbook methods (characteristics, Froebnius, Fourier transform) fail.   Does any one know weather a method exists to solve this?  Apologies if this is a simple question but, well, I am a theoretical physicist and it is not simple for me.
 A: Hint : a way to reduced form of PDE
$$
y \left(\alpha \frac{\partial }{\partial y}+x \frac{\partial^2 }{\partial x \partial y} \right)f(x,y) = \left( z_1 + z_2 x^\alpha y^{-2} \right) f(x,y)$$
$  \begin{cases}
    y=e^Y      \\
    x=e^{\frac{X}{\alpha}}  \\
    f(x,y)=g(X,Y) \\
  \end{cases}$
$\quad $
$  \begin{cases}
    dy=ydY      \\
    dx=\frac{1}{\alpha}xdX  
  \end{cases}$
$\quad $
$\alpha \frac{\partial g}{\partial Y}+\alpha \frac{\partial^2 g}{\partial X \partial Y}=\left( z_1+z_2e^{X-2Y}\right) g$ 
$  \begin{cases}
    z_1=\alpha c_1  \\
    z_2=\alpha c_2
  \end{cases}$
$\quad \quad\quad  $
$\frac{\partial g}{\partial Y}+\frac{\partial^2 g}{\partial X \partial Y}=
\left( c_1+c_2e^{X-2Y}\right)g$
This is a kind of hyperbolic PDE on the form :
$$\frac{\partial^2 g}{\partial X \partial Y}+\frac{\partial g}{\partial Y}+
F(X,Y)g(X,Y)=0$$
where $F(X,Y)=\left( c_1+c_2e^{X-2Y}\right)$
Hopping that could help...
A: Following up on JJacquelin's answer and not guaranteeing that no mistakes
are made.
We have
\begin{equation*}
\partial _{Y}(1+\partial _{X})g+(c_{1}+c_{2}e^{X-2Y})g
\end{equation*}
If you put $u=X-2Y$ \ then
\begin{equation*}
\partial _{Y}(1+\partial _{u})g+(c_{1}+c_{2}e^{u})g=0
\end{equation*}
Now let $h=e^{u}g$ then
\begin{equation*}
(1+\partial _{u})e^{-u}h=e^{-u}\partial _{u}h
\end{equation*}
\begin{equation*}
\partial _{Y}\partial _{u}h=-(c_{1}+c_{2}e^{u})h
\end{equation*}
Next put $h=m(Y)n(u)$. Then
\begin{eqnarray*}
\partial _{Y}\ln m(Y)\partial _{u}\ln n(u) &=&-(c_{1}+c_{2}e^{u}) \\
\partial _{Y}\ln m(Y) &=&-\frac{(c_{1}+c_{2}e^{u})}{\partial _{u}\ln n(u)}=M
\end{eqnarray*}
\begin{eqnarray*}
m(Y) &=&\exp [MY] \\
\partial _{u}\ln n(u) &=&-M^{-1}(c_{1}+c_{2}e^{u}) \\
n(u) &=&\exp [-M^{-1}(c_{1}u+c_{2}e^{u})]
\end{eqnarray*}
\begin{eqnarray*}
g &=&e^{-u}\exp [MY]\exp [-M^{-1}(c_{1}u+c_{2}e^{u})] \\
u &=&X-2Y
\end{eqnarray*}
Edit: There were some mistakes. Below a new attempt.
With
\begin{equation*}
h(X,Y)=e^{X}g(X,Y)
\end{equation*}
we have
\begin{equation*}
\partial _{Y}\partial _{X}h(X,Y)=(c_{1}+c_{2}e^{X-2Y})h(X,Y)
\end{equation*}
New variables
\begin{equation*}
U=X-2Y,\;V=X+2Y
\end{equation*}
Then ($\left. {}\right\vert _{W}$ is the partial derivative with $W$
kept constant)
\begin{eqnarray*}
\left. \partial _{X}\right\vert _{Y} &=&\left. \partial _{U}\right\vert
_{V}+\left. \partial _{V}\right\vert _{U},\;\left. \partial _{Y}\right\vert
_{X}=-2\{\left. \partial _{U}\right\vert _{V}-\left. \partial
_{V}\right\vert _{U}\} \\
(\left. \partial _{X}\right\vert _{Y})\left. \partial _{Y}\right\vert _{X}
&=&2\{\left. \partial _{V}\right\vert _{U}^{2}-\left. \partial
_{U}\right\vert _{V}^{2}\}
\end{eqnarray*}
With
\begin{equation*}
h(X,Y)=k(U,V)
\end{equation*}
we then have
\begin{equation*}
2\{\left. \partial _{V}\right\vert _{U}^{2}-\left. \partial _{U}\right\vert
_{V}^{2}\}k(U,V)=(c_{1}+c_{2}e^{U})k(U,V)
\end{equation*}
Now try
\begin{equation*}
k(U,V)=m(U)n(V)
\end{equation*}
Then, omitting the subscripts on the derivatives,
\begin{eqnarray*}
m(U)\partial _{V}^{2}n(V)-\partial _{U}^{2}m(U)n(V) &=&\frac{1}{2}%
(c_{1}+c_{2}e^{U})m(U)n(V) \\
\frac{\partial _{V}^{2}n(V)}{n(V)}-\frac{\partial _{U}^{2}m(U)}{m(U)} &=&
\frac{1}{2}(c_{1}+c_{2}e^{U}) \\
\frac{\partial _{V}^{2}n(V)}{n(V)} &=&\frac{\partial _{U}^{2}m(U)}{m(U)}+
\frac{1}{2}(c_{1}+c_{2}e^{U})=M
\end{eqnarray*}
where $M$ is a separation constant. Hence
\begin{eqnarray*}
\partial _{V}^{2}n(V) &=&Mn(V) \\
\partial _{U}^{2}m(U) &=&\{M-\frac{1}{2}(c_{1}+c_{2}e^{U})\}m(U)
\end{eqnarray*}
