Consider the PDE
$$
(1+x^2)^2u_{xx}-u_{yy}+2x(1+x^2)u_x=0\text{ in } \mathbb{R}^2.
$$
Transform the PDE in normal form (with mixted derivative). And find a general solution of the normal form and the original PDE.
Hello, everybody!
It is a rather long calculation, therefore I only give you my result for the normal form, if it is okay. But maybe I should at least say which transformation I used:
$$
\xi:=y-\arctan(x),~~~~~\eta:=y+\arctan(x).
$$
I get the very simple normal form with mixted dervative
$$
v_{\xi\eta}=0
$$
Is that right?
If yes: I determined the general solution of it as
$$
v(\xi,\eta)=\int_{\eta_0}^{\eta}f(\tau)\, d\tau+g(\xi),
$$
whereat $f\in C^1(\mathbb{R}), g\in C^2(\mathbb{R})$ and $\eta_0\in\mathbb{R}$ arbitrary.
Hope, that's right. Then the original PDE has the general solution
$$
u(x,y)=\int_{z_0}^{y+\arctan(x)}f(\tau)\, d\tau+g(y-\arctan(x)).
$$
Best Answer
Your calculations are correct. The general solution of $u_{\xi\eta}=0$ (in terms of $x,y$) is $u(x, y)=F(y+\arctan x)+G(y-\arctan x)$, where $F, G$ are arbitrary $C^2$ functions.