# Charpit's Method

Find the complete integral of partial differential equation
z^2 = pqxy ?
I have solved this equation till auxiliary equation:

dp/(-pqy+2pz) = dq/(-pqx+2qz) = dz/(2pqxy) = dx/(qxy) = 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

-
Can you make clear what the question is? I see an equation with five variables, not a differential equation at all. What is a function of what, and where are the differentials? From the third line, maybe z, p, and q are all functions of x and y, but unless some are specified there is not enough information for a solution. –  Ross Millikan Dec 6 '10 at 14:13
So is the equation $z^2=xy\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}$ where z is a function of two variables? –  Ross Millikan Dec 6 '10 at 16:52

If the equation is $z^2=xy\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}$, I would be tempted to see the symmetry in $x$ and $y$ and try solutions of the form $z=(xy)^n$. What happens then?

-
What happens is that $n=\pm1$. And then? –  Did Dec 14 '11 at 10:22
So you have a solution $z-xy=0, F_x=p=y, F_y=q=x$ But F_p seems to be $0$ for Charpit's method. –  Ross Millikan Dec 14 '11 at 14:12
Right. I asked because I thought we were after every solution of this pde, but rereading the question I am not so sure anymore... –  Did Dec 14 '11 at 14:31
More generally, you can have $z=c^2(xy)^c$ or $z=-c^2(\frac xy)^c$ for $c$ positive or negative. –  Ross Millikan Dec 14 '11 at 16:25