$\displaystyle 2uq^2+y^2(q-p)=0$

$\displaystyle p= \partial u/\partial x $; $\displaystyle q= \partial u/\partial y $

We get the charpit equations:

$\displaystyle \frac{dx}{-y^2}=\frac{dy}{4uq+y^2}=\frac{du}{-py^2+q(4uq+y^2)}=\frac{dp}{-2pq^2}=\frac{dq}{-2y(q-p)-2q^3}$

I don't know how to find the value of p and q ??

(i have tried the following ways but i cannot:

from the original equation we have: $\displaystyle (q-p) =-\frac{2uq^2}{y^2} $

substitute into the last pair equation $\displaystyle \frac{dq}{-2y(-\frac{2uq^2}{y^2})-2q^3}=\frac{dp}{-2pq^2}$

or $\displaystyle \frac{dq}{dp}=\frac{1}{p}(\frac{-2u}{y}+q) => q = \frac{2u}{y}+p.c1$

but i don't know how to find p? because any pairs will lead to a complicated equation. please experts help!! thanks )