<s>[INST] Solve a PDE: $ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$ Solve the PDE $$ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$$
where, $ p=\displaystyle \frac{\partial z}{\partial x}$ and $ q=\displaystyle \frac{\partial z}{\partial y}$
My attempt:
I start with Lagrange's auxiliary equation
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{z(x^2-y^2)}$$
Relation 1: $$\frac{x \, dx+y \, dy- dz}{x^2y^2+x^2z-y^2x^2-y^2z-x^2z+y^2z} = \frac{x \, dx+y \, dy-dz}{0} $$
Integrating $x \, dx+y \, dy-dz=0$
$$\frac{x^2}{2}+\frac{y^2}{2}-z=c \implies x^2+y^2-2z=c_1 $$
I can't see/find a second relation for me to solve the problem. Please help.
 [/INST] since
$$\dfrac{\dfrac{dx}{x}}{y^2+z}=\dfrac{\dfrac{dy}{y}}{-x^2-z}=\dfrac{dz}{x^2-y^2}$$
$$\Longrightarrow \dfrac{\dfrac{dx}{x}+\dfrac{dy}{y}}{y^2-x^2}=\dfrac{dz}{x^2-y^2}$$
$$\Longrightarrow \dfrac{dx}{x}+\dfrac{dy}{y}=-dz$$
$$\Longrightarrow z=-C\ln|xy|$$
</s>