Charpit's Method Find the complete integral of partial differential equation
$$\displaystyle z^2 = pqxy $$ 
I have solved this equation till auxiliary equation:
$$\displaystyle \frac{dp}{-pqy+2pz}=\frac{dq}{-pqx+2qz}=\frac{dz}{2pqxy}=\frac{dx}{qxy}=\frac{dy}{pxy} $$  
But I have unable to find value of p and q.
EDIT:
p = ∂z/∂x
q = ∂z/∂y
r = ∂²z/∂x²  = ∂p/∂x
s = ∂²z/∂x∂y = ∂p/∂y or ∂q/∂x
t = ∂²z/∂y²  = ∂q/∂y


A much easier solution can be obtained by introducing new dependent/independent variables U=log u, X=log x, Y=log y. Then, with P,Q denoting the first partial derivatives of U with respect to X,Y, respectively, the PDE becomes
PQ=1,
which can be solved very easily by Charpit's method.
