How can I solve the second order PDE? How can I solve the second order PDE?
$x^2u_{xx}+2xyu_{xy}+y^2u_{yy}+xyu_x+y^2u_y=0$.
Solution:
I find $\Delta=0$.Then, $\frac{dy}{dx}=\frac{y}{x}$ and from here I find $\frac{x}{y}=c_1$
Now, will I define $\xi(x,y)=\frac{x}{y}$ and $\eta=y$ ? Can you help me?
 A: Of course, nothing prevents you from taking $\eta = y$ in this parabolic equation. You must make sure that the transformation $(x,y) \to (\xi,\eta)$ is not 'degenerate' or 'singular' (sorry if I'm not wording this accurately), like, for instance, $\eta = 1$ (you would be mapping variables into a constant!). 
But it's good practice to convince yourself that this is indeed a good choice for $\eta$. Indeed, if you let $\eta = f(x,y)$ for some general $f$ and work out the partial derivatives in terms of $\xi$ and $\eta$ you'll find:
$$ (x f_x + y f_y)^2 \, u_{\eta \eta} + (x^2 f_{xx} + y^2 f_{yy} + xy f_x + y^2 f_y) \, u_\eta = 0 $$
If, as you suggest, one takes the good choice $\eta = y$, the following 'ODE' for $u$ is found:
$$ u_{\eta \eta} + u_\eta = 0$$
Solving by means of any of your favourite methods, you have
$$u(\xi,\eta)  = A(\xi) + B(\xi) \mathrm{e}^{-\eta}$$
where $A$ and $B$ are arbitrary functions of their arguments. As you can see, everything turned out well and easy. I leave it to you to think what would happen for another choice of $\eta$. 
Moreover, what happens for $y=0$? And for $x = 0$? And for $x=y$?
