Solving a PDE through separation of variables I have the following PDE: 
$$x^2u_{xx} - y^2u_{yy}-2yu_y = 0 .$$
after seperating variables, I obtain after separating variables, I obtain $$\frac{x^2}{\phi} \phi '' = - \lambda ,$$ and  $$\frac{y^2}{g} g ''  -\frac{2y}{g} g' = -\lambda,$$  where $u(x,y) = \phi(x)g(y)$ and $\lambda$ is the separation constant. How should I proceed? 
 A: It is more powerful to use the method of characteristics (MoC) than separation of variables (SoV) for OP's problem. In this answer we give a sketched derivation of the full solution using the MoC. 


*

*Notice that OP's linear 2nd order PDE 
$$x^2 u_{xx}-y^2 u_{yy}-2yu_y~=~0 $$
is form-invariant under scalings $(x,y)\to (\lambda z, \mu y)$.
This suggests that we should define new variables
$$X~=~\ln|x|\quad\text{and}\quad Y~=~\ln|y|.$$
Then OP's linear 2nd order PDE becomes a linear 2nd order PDE 
$$u_{XX}-u_X-u_{YY}-u_Y~=~0  $$
with constant coefficient functions.

*If we define 
$$ X^{\pm}~=~X\pm Y, $$
the PDE factorizes
$$ (4\partial_--2)\partial_+u~=~0, $$
or equivalently,
$$ \partial_- e^{-X^-/2}\partial_+u~=~0, $$
with full solution
$$u~=~ F(X^-)+ e^{X^-/2}G(X^+),$$
where $F,G$ are arbitrary functions.

*Going back to the original variables, the solution reads
$$ u~=~ f(x/y) + \sqrt{\left| \frac{x}{y}\right|} g(xy), $$
where $f,g$ are arbitrary functions. This can in turn be beautified to$^1$
$$ u~=~ f(x/y) + x g(xy)  $$
for arbitrary functions $f,g$. The above full solution would have been more difficult to derive using SoV.
--
$^1$ Hat tip: Robert Israel.
