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: 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?
A: let the given equation be 
$$f(z,p,q)=0  \text{ i.e, } z^2-pqxy=0$$
perform derivation w.r.t p,q,x,y,z
then write charpits relation 
$$\frac {dx}{-fp} =\frac {dy}{-fq} =\frac {dz}{(-p*fp-q*fq)} =\frac {dp}{(fx+x*fz)} =\frac {dq}{(fy+y*fz)}$$
where $fp,fq,fx,fy,fz$ are derivatives of $z^2-pqxy$ I think u got it :)
after substitutions
equate 1 and three equations i.e 
$$ \frac {dx}{qxy}=\frac {dz}{2pqxy}$$
now u can find the value of 
$$p=\frac {z+c}{2x} $$
put p in 
$$z^2-pqxy=0$$
 u can find 
$$q=\frac  {2z^2}{((z+c)y)}$$
put the p,q in 
$$dz=pdx+qdy$$
 and integrate u can get the solution :) if im wrong please correct me thanking u
A: Here is your solution.
perform 
$$\frac {p dp - q dq}{{pq(qy-px)}} = - \frac {ydx-xdy}{{xy(px-qy)}} $$
resulting to 
$$\frac {d(pq)}{pq}=\frac {d(xy)}{xy}$$
Integrating we get 
$$\log pq = \log xy+ \log c  \implies \frac {pq}{xy}=c \implies p= \frac {cxy}q. $$
Substitute this value in your prob. and proceed as usual.
A: using multipliers p,q,x & y in 1st, 2nd, 4th & 5th equations and equating it with equation 3rd.
$$\implies  \frac {dz}{2pqxy}=(pdx+qdy+xdp+ydq)/pqxy+pqxy-pqxy+2pxz-pqxy+2qyz$$
from question $z^2=pqxy$:
$$\implies dz/(2z^2)= (pdx+qdy+xdp+ydq)/2pzx+2qxy$$
$$ \implies dz/2(z^2)={d(px)+d(qy)}/2z(px+qy) $$
$$\implies \frac {dz}z=\frac {d(px+qy)}{px+qy}$$
$$\implies \ln(z)=\ln(px+qy)+\ln(a)$$
$$ \implies z=a(px+qy).$$
A: from your auxiliary equations: use
$$\frac{zp\ dx + xz\ dp}{xyzpq - xyzpq +2pxz^2} = \frac{q\ dy + y\ dq}{pqxy - pqxy + 2yzq}. $$
$$\implies \frac{p\ dx + x\ dp}{2pxz} = \frac{d(yq)}{2yzq} \implies \frac{d(xp)}{xp} = \frac{d(yq)}{yq}$$. on integrating we get 
$xp = yqa$ ($a$=constant). then you complete using  $dz=p\ dx+q\ dy$  (using the given equation $z^2 = pqxy$)
A: 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.
