In my work I have faced with following partial differential equation
$$\left(\frac{\partial u}{\partial x}\right)^2-\left(\frac{\partial u}{\partial y}\right)^2+f(x,y)\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}=0$$
where $f(x,y)=\frac{1}{2}\frac{(y^2-x^2)^2}{xy(x^2+y^2)}$, $u=u(x,y)$
I've tried to solve it by method of characteristic, but it was too complicated.
Maybe someone knows something about this type of PDE? Anyway I would be grateful for any hints.
Hint:
$\left(\dfrac{\partial u}{\partial x}\right)^2-\left(\dfrac{\partial u}{\partial y}\right)^2+\dfrac{1}{2}\dfrac{(y^2-x^2)^2}{xy(x^2+y^2)}\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial y}=0$
$2xy(x^2+y^2)\left(\dfrac{\partial u}{\partial x}\right)^2+(x^2-y^2)^2\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial y}-2xy(x^2+y^2)\left(\dfrac{\partial u}{\partial y}\right)^2=0$
$\dfrac{\partial u}{\partial x}=\dfrac{-(x^2-y^2)^2\dfrac{\partial u}{\partial y}\pm\sqrt{(x^2-y^2)^4\left(\dfrac{\partial u}{\partial y}\right)^2+16x^2y^2(x^2+y^2)^2\left(\dfrac{\partial u}{\partial y}\right)^2}}{4xy(x^2+y^2)}$
$\dfrac{\partial u}{\partial x}=\dfrac{-(x^4-2x^2y^2+y^4)\pm(x^4+6x^2y^2+y^4)}{4xy(x^2+y^2)}\dfrac{\partial u}{\partial y}$ (according to http://www.wolframalpha.com/input/?i=factorize%28x%5E2-y%5E2%29%5E4%2B16x%5E2y%5E2%28x%5E2%2By%5E2%29%5E2)
$\dfrac{\partial u}{\partial x}=\dfrac{2xy}{x^2+y^2}\dfrac{\partial u}{\partial y}$ or $\dfrac{\partial u}{\partial x}=-\dfrac{x^2+y^2}{2xy}\dfrac{\partial u}{\partial y}$