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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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    $\begingroup$ 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. $\endgroup$ Dec 6, 2010 at 14:13
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    $\begingroup$ 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? $\endgroup$ Dec 6, 2010 at 16:52

6 Answers 6

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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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  • $\begingroup$ What happens is that $n=\pm1$. And then? $\endgroup$
    – Did
    Dec 14, 2011 at 10:22
  • $\begingroup$ 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. $\endgroup$ Dec 14, 2011 at 14:12
  • $\begingroup$ Right. I asked because I thought we were after every solution of this pde, but rereading the question I am not so sure anymore... $\endgroup$
    – Did
    Dec 14, 2011 at 14:31
  • $\begingroup$ More generally, you can have $z=c^2(xy)^c$ or $z=-c^2(\frac xy)^c$ for $c$ positive or negative. $\endgroup$ Dec 14, 2011 at 16:25
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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

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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 $$\frac {p dp - q dq}{{pq(qy-px)}} = - \frac {ydx-xdy}{{xy(px-qy)}} $$

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.

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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 $z^2=pqxy$:

$$\implies dz/(2z^2)= (pdx+qdy+xdp+ydq)/2pzx+2qxy$$ $$ \implies dz/2(z^2)={d(px)+d(qy)}/2z(px+qy) $$ $$\implies \frac {dz}z=\frac {d(px+qy)}{px+qy}$$ $$\implies \ln(z)=\ln(px+qy)+\ln(a)$$ $$ \implies z=a(px+qy).$$

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    $\begingroup$ You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. $\endgroup$ Feb 25, 2013 at 7:28
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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.

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