Solution of the following nonlinear PDE Find general solution of the PDE
$$x^2u_x^2+y^2u_y^2=u^2$$
Solution:
when we use transform $v=\ln u$ (or $u=e^v$), we get $x^2v_x^2+y^2v_y^2=1$. 
And by the using separation of variables $v(x,y)=f(x)+g(y)$, I got the $x^2f'^2=\lambda^2$ and $g'^2=(1-\lambda^2)/y^2$.
And how can we solve the rest?
 A: Hint:
Let $u=e^v$ ,
Then $u_x=e^vv_x$
$u_y=e^vv_y$
$\therefore x^2e^{2v}v_x^2+y^2e^{2v}v_y^2=e^{2v}$
$x^2v_x^2+y^2v_y^2=1$
Let $\begin{cases}p=\ln x\\q=\ln y\end{cases}$ ,
Then $v_x=v_pp_x+v_qq_x=\dfrac{v_p}{x}$
$v_y=v_pp_y+v_qq_y=\dfrac{v_q}{y}$
$\therefore x^2\dfrac{v_p^2}{x^2}+y^2\dfrac{v_q^2}{y^2}=1$
$v_p^2+v_q^2=1$
A: Why not use separation of variables from the beginning. Introduce $u=X(x)Y(y)$ to get
$$x^2X'^2Y^2+y^2X^2Y'^2=X^2Y^2 \implies \frac{x^2X'^2}{X^2}+\frac{y^2Y'^2}{Y^2}=1$$
It is clear that the equation can only be true if $\frac{x^2X'^2}{X^2}$ and $\frac{y^2Y'^2}{Y^2}$ are constants such that their sum adds up to 1. So we obtain two ordinary differential equations:
$$X'(x)^2=k^2\frac{X^2(x)}{x^2}$$
$$Y'(y)^2=(1-k^2)\frac{Y(y)^2}{y^2}$$
Now we take the square root of both expressions
$$X'(x)=\pm k\frac{X(x)}{x} \implies X(x)=c_1x^{\pm k}$$
$$Y'(y)=\pm \sqrt{1-k^2}\frac{Y(y)}{y} \implies Y(y)=c_2y^{\pm \sqrt{1-k^2}}.$$
Now, the solution by separation of variables is given by
$$u(x,y)=c_1c_2x^{\pm k}y^{\pm \sqrt{1-k^2}}.$$
Note, that you cannot superimpose different solutions for $k$ as the differential equation is not linear.
EDIT: If you want to solve 
$$f'(x)^2=\frac{\lambda^2}{x^{2}}\implies f'(x)=\pm\frac{\lambda}{x} \implies f(x)=\pm \lambda\ln x +c_1$$
$$g'(y)^2=\frac{1-\lambda^2}{y^2} \implies g'(y)=\pm\frac{\sqrt{1-\lambda^2}}{y} \implies g(y)=\pm\sqrt{1-\lambda^2}\ln y +c_2$$
You will see that $|\lambda|\leq 1$ is a side constraint such that the solutions stay real.
