Charpit's Method

Question

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

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 06 '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 06 '10 at 16:52

score 5 · Accepted Answer · edited Apr 12 '18 at 16:18

5

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?

edited Apr 12 '18 at 16:18

user577215664

40,625

answered Dec 06 '10 at 19:18

Ross Millikan

374,822

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

score 2 · Answer 2 · edited Apr 12 '18 at 16:19

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

score 1 · Answer 3 · edited Apr 07 '15 at 19:50

1

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 $z^2 = pqxy$)

edited Apr 07 '15 at 19:50

jameselmore

5,207

answered Apr 07 '15 at 19:22

salihu sabiu musa

11

score 1 · Answer 4 · edited Jan 07 '21 at 12:12

1

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 $z^2=pqxy$:

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

edited Jan 07 '21 at 12:12

amWhy

209,954

answered Feb 25 '13 at 07:23

Roshan

11

3

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. – Zev Chonoles Feb 25 '13 at 07:28

score 1 · Answer 5 · edited Apr 12 '18 at 16:16

1

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.

edited Apr 12 '18 at 16:16

user577215664

40,625

answered Dec 14 '11 at 09:37

Sfurti A. Verma

11

2

For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Nov 04 '12 at 12:00

score 0 · Answer 6 · answered Dec 08 '22 at 13:22

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.

Charpit's Method

6 Answers6