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I've been attempting to solve this non-linear PDE

$$4\Omega x^2 y^2 \frac{\partial z}{\partial y} -x^2 y (\frac{\partial z}{\partial x})^2 + 2x^2 y^2 E-N^2=0$$

using Charpit's method. The variables $\Omega$, $E$, and $N$ are constants. I've derived the relation between Charpit's auxillary equations,

$$\frac{x}{N^2}dp=-\frac{y}{N^2}dq=-\frac{1}{x^2 y p}dx=\frac{1}{2\Omega x^2 y^2} dy$$

but I have been unable to separate these to obtain the second function relating $p$ and $q$. Most of the examples I have seen are separated very easily with terms such as

$$\frac{dp}{p}=\frac{dq}{q}$$

or something similar. Could anyone please help?

EDIT: A little background on Charpit's method and the Method of Characteristics...

We begin by defining the primary non-linear PDE

$$F(x,y,z,p,q)=4\Omega x^2 y^2 \frac{\partial z}{\partial y} -x^2 y (\frac{\partial z}{\partial x})^2 + 2x^2 y^2 E-N^2=0$$

Then, by expanding the total derivatives

$$\frac{dF}{dx}=0$$ and $$\frac{dF}{dy}=0$$

in terms of partial derivatives, along with the definitions

$$\frac{dx}{dt}\equiv\frac{\partial F}{\partial p}$$ and $$\frac{dy}{dt}\equiv\frac{\partial F}{\partial q}$$

we get the five Charpit Equations:

$$\frac{dx}{dt}=\frac{\partial F}{\partial p}\\ \frac{dy}{dt}=\frac{\partial F}{\partial q}\\ \frac{dp}{dt}=-\frac{\partial F}{\partial x}-p\frac{\partial F}{\partial z}\\ \frac{dq}{dt}=-\frac{\partial F}{\partial y}-q\frac{\partial F}{\partial z}\\ \frac{dz}{dt}=p\frac{dx}{dt}+q\frac{dy}{dt}$$

By eliminating $dt$ they can all be set equal to each other (see Charpit's auxillary equations mentioned above), and any two (or more) can be used to integrate a total derivative relating $p$ and $q$. This new relation is then substituted into the original differential equation, which can then be written as

$$dz=p(x,y)dx+q(x,y)dy$$

which is a total differential which can be directly integrated to find the solution z=z(x,y).

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  • $\begingroup$ Providing a link to Charpit's method might help... $\endgroup$ – Zach466920 Apr 30 '15 at 22:29
  • $\begingroup$ Of course! Here's a good resource I've used (Google Books): tinyurl.com/la57y4u $\endgroup$ – William J. Cunningham May 1 '15 at 12:41
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Hint:

$4\Omega x^2y^2\dfrac{\partial z}{\partial y}-x^2y\left(\dfrac{\partial z}{\partial x}\right)^2+2x^2y^2E-N^2=0$

$4\Omega y\dfrac{\partial z}{\partial y}-\left(\dfrac{\partial z}{\partial x}\right)^2+2Ey-\dfrac{N^2}{x^2y}=0$

$4\Omega y\dfrac{\partial^2z}{\partial x\partial y}-2\dfrac{\partial z}{\partial x}\dfrac{\partial^2z}{\partial x^2}+\dfrac{2N^2}{x^3y}=0$

$\dfrac{\partial z}{\partial x}\dfrac{\partial^2z}{\partial x^2}-2\Omega y\dfrac{\partial^2z}{\partial x\partial y}=\dfrac{N^2}{x^3y}$

Let $u=\dfrac{\partial z}{\partial x}$ ,

Then $u\dfrac{\partial u}{\partial x}-2\Omega y\dfrac{\partial u}{\partial y}=\dfrac{N^2}{x^3y}$

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  • $\begingroup$ I understand that this is now quasi-linear instead of fully non-linear but I'm still having trouble solving it :/ $\endgroup$ – William J. Cunningham May 7 '15 at 14:42

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