Solve a PDE: $ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$

Question

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.

Answer 1

since $$\dfrac{\dfrac{dx}{x}}{y^2+z}=\dfrac{\dfrac{dy}{y}}{-x^2-z}=\dfrac{dz}{x^2-y^2}$$ $$\Longrightarrow \dfrac{\dfrac{dx}{x}+\dfrac{dy}{y}}{y^2-x^2}=\dfrac{dz}{x^2-y^2}$$ $$\Longrightarrow \dfrac{dx}{x}+\dfrac{dy}{y}=-dz$$ $$\Longrightarrow z=-C\ln|xy|$$

Answer 2

Using multipliers $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$,

Relation 2: $$\frac{\frac{1}{x} \, dx+\frac{1}{y} \, dy +\frac{1}{z} dz}{y^2+z-x^2-z+x^2-y^2} = \frac{\frac{1}{x} \, dx+\frac{1}{y} \, dy +\frac{1}{z} dz}{0} $$

Integrating $\frac{1}{x} \, dx+\frac{1}{y} \, dy +\frac{1}{z} dz=0$

$$\log x+\log y+\log z=c \implies xyz=e^c = c_2 $$