How to solve the following pde? How to solve the following PDE? 
For an arbitrary continuously differentiable function $f$ , which of the following is a general solution of $\;$
$z(px-qy)=y^2-x^2$?
1)$\;$ $x^2+y^2+z^2=f(xy)$
2)$\;$  $(x+y)^2+z^2=f(xy)$
3)$\;$ $x^2+y^2+z^2=f(y-x)$
4)$\;$ $x^2+y^2+z^2=f((x+y)^2+z^2)$
Here options $1)$$\;$ $2)$  and $ 4)$ are correct
but I am getting only first option as answer.
Is there any general method to solve such pdes?
Please help because I don't have any teacher who can help me and I am learing pde without any teacher.
Thank you very much for giving me your precious time.
 A: $z(xz_x-yz_y)=y^2-x^2$
$xz_x-yz_y=\dfrac{y^2-x^2}{z}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$
$\dfrac{dz}{dt}=\dfrac{y^2-x^2}{z}=\dfrac{y_0^2e^{-2t}-e^{2t}}{z}$ , we have $z^2=f(y_0)-y_0^2e^{-2t}-e^{2t}=f(xy)-y^2-x^2$ , i.e. $x^2+y^2+z^2=f(xy)$
$\therefore$ 1) is obviously correct.
But 2) is in fact also correct since $(x+y)^2=x^2+2xy+y^2$ .
From 2), $g((x+y)^2+z^2)=xy$
$\therefore x^2+y^2+z^2=f(g((x+y)^2+z^2))$ , equivalent to $x^2+y^2+z^2=f((x+y)^2+z^2)$
Hence 4) is in fact also correct.
See rule 6 for further details.
A: Adding another way of looking at it: Taking Lagrange Multiplier,
$${dx\over zx}={dy\over -zq}={dz\over y^2-x^2}$$
First two gives $xy=c$ and $x\times$ first $+y\times$second=$-z\times$third gives
$${(x+y)d(x+y)\over zx^2-zy^2}={-zdz\over zx^2-zy^2}$$
and gives $(x+y)^2+z^2=c$.
That adds (2) and (4)
