Solve the PDE : $z(p^2-q^2)=x-y$ We have to find complete integral.
I am finding difficulty in converting it into a equation not involving $z$.
I tried solving it using charpit method but I got stuck.
 A: Hint:
Let $\begin{cases}u=x+y\\v=x-y\end{cases}$ ,
Then $\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}$
$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=\dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v}$
$\therefore z\left(\dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}\right)\left(\dfrac{\partial z}{\partial x}-\dfrac{\partial z}{\partial y}\right)=x-y$
$4z\dfrac{\partial z}{\partial u}\dfrac{\partial z}{\partial v}=v$
Let $z=w^\frac{2}{3}$ ,
Then $\dfrac{\partial z}{\partial u}=\dfrac{\partial z}{\partial w}\dfrac{\partial w}{\partial u}=\dfrac{2}{3w^\frac{1}{3}}\dfrac{\partial w}{\partial u}$
$\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial w}\dfrac{\partial w}{\partial v}=\dfrac{2}{3w^\frac{1}{3}}\dfrac{\partial w}{\partial v}$
$\therefore4w^\frac{2}{3}\dfrac{2}{3w^\frac{1}{3}}\dfrac{\partial w}{\partial u}\dfrac{2}{3w^\frac{1}{3}}\dfrac{\partial w}{\partial v}=v$
$\dfrac{\partial w}{\partial u}\dfrac{\partial w}{\partial v}=\dfrac{9v}{16}$
It belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3318.pdf.
