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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
  • 1
    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.2010-12-06
  • 1
    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?2010-12-06

5 Answers 5

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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?

  • 0
    What happens is that $n=\pm1$. And then?2011-12-14
  • 0
    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.2011-12-14
  • 0
    Right. I asked because I thought we were after *every* solution of this pde, but rereading the question I am not so sure anymore...2011-12-14
  • 0
    More generally, you can have $z=c^2(xy)^c$ or $z=-c^2(\frac xy)^c$ for $c$ positive or negative.2011-12-14
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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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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.

  • 2
    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-11-04
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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 $z^2=pqxy$

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

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    You can find some good starting points on how to format mathematics on the site [here](http://meta.math.stackexchange.com/questions/5020). [This AMS reference](ftp://ftp.ams.org/ams/doc/amsmath/short-math-guide.pdf) 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](http://tex.stackexchange.com/) site.2013-02-25
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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 $z^2 = pqxy$)