PDE - $y^2 \frac{\partial^2 u}{\partial x^2}=x^2 \frac{\partial^2 u}{\partial y^2}$ - how to derive the general solution $\mathbf{y^2 \frac{\partial^2 u}{\partial x^2}=x^2 \frac{\partial^2 u}{\partial y^2}}$  
is a hyperbolic PDE where   
$\xi =y^2+x^2$  $\eta =y^2-x^2$ 
which gives 
$u_{xx}=2(u_{\xi}-u_{\eta})+4x^2(u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta})$ 
$u_{yy}=2(u_{\xi}+u_{\eta})+4y^2(u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta})$ 

$y^2=\frac{1}{2}(\xi+\eta)$ 
$x^2=\frac{1}{2}(\xi-\eta)$ 
$4x^2 y^2=\xi^2-\eta^2$ 
so I got the below canonical form which I don't know how to convert into a general solution
$\mathbf{-4(\xi^2-\eta^2)u_{\xi\eta}=0}$
can anyone explain further steps?
 A: Here's what I did, I hope this is helpful:
Rewrite this as: $(\frac{y}{x})^2 \frac{\partial^2 u}{\partial x^2} = \frac{ \partial^2 u}{ \partial y^2}$. Introduce the parameter $\alpha= \frac{y}{x}$. Our goal is to rewrite the PDE purely in terms of alpha. Luckily, this can be done and we obtain the following (the details I'll leave to you, they mostly consist of repeated application of the chain rule):
$(\alpha^4-1) \frac{\partial^2 u}{\partial \alpha^2} + 2 \alpha^3 \frac{\partial u}{\partial \alpha}=0$
Note that we can write this as:
$\frac{u_{\alpha \alpha}}{u_{\alpha}}= \frac{-2 \alpha^3}{\alpha^4-1}$ (using subscript notation for partials). Which is equivalent to:
$\frac{\partial}{\partial \alpha} \ln (u_{\alpha})= \frac{-2 \alpha^3}{\alpha^4-1}$
Integrating:
$\ln (u_{\alpha})= -\frac{1}{2} \ln (\frac{1}{4} (\alpha^4-1)) +C$
Now if we assume $u$ can be written solely as a function of the parameter $\alpha$ (in which case we've effectively transformed this into an ODE), then $C$ is a fixed constant. Else, $C$ may be an arbitrary function, i.e. $C=C(x,y)$. We get an expression for $u_\alpha$:
$u_\alpha= \frac{2}{\sqrt{\alpha^4-1}}+C$
Now integrating both sides doesn't evaluate nicely, i.e. according to Wolfram Alpha, there is no representation in terms of elementary functions. Furthermore, to evaluate the $C$ component through integration, you either need to assume $C$ is in fact a constant or you need sufficient boundary conditions. 
