# 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

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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?

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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

let the given equation be f(z,p,q)=0 i.e, z^2-pqxy=0 perform derivation w.r.t p,q,x,y,z then write charpits relation dx/-fp =dy/-fq =dz/(-p*fp-q*fq) =dp/(fx+x*fz) =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 dx/(qxy)=dz/(2pqxy) now u can find the value of p=(z+c)/(2*x) put p in z^2-pqxy=0 u can find q=2*z^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

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using multipliers p,q,x & y in 1st, 2nd, 4th & 5th equations and equating it with equation 3rd. => dz/(2pqxy)=(pdx+qdy+xdp+ydq)/pqxy+pqxy-pqxy+2pxz-pqxy+2qyz from question z^2=pqxy => dz/(2z^2)= (pdx+qdy+xdp+ydq)/2pzx+2qxy =>dz/2(z^2)={d(px)+d(qy)}/2z(px+qy) =>dz/z={d(px+qy)}/px+qy =>ln(z)=ln(px+qy)+ln(a) =>z=a(px+qy)

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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 $z^2 = pqxy$)

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Here is your solution. perform $(p dp - q dq)/{pq(qy-px)} = - (ydx-xdy)/{xy(px-qy)}$resulting to $d(pq)/pq=d(xy)/xy$ Integrating we get $log pq = log xy+ log c \implies (pq)/(xy)=c \implies p= cxy/q.$Substitute this value in your prob. and proceed as usual.

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