Finding prime integrals for a PDE

Question

Hello I started learning about PDE's recently, and I find quite difficult finding prime integrals, could you help me with this one? $$2xz\frac{\partial z}{\partial x}+2yz\frac{\partial z}{\partial y}=z^2-x^2-y^2$$ I have written the characteristic sysyem: $$\frac{dx}{2xz}=\frac{dy}{2yz}=\frac{dz}{z^2-x^2-y^2}$$ Now, I multiply by $z$ the first two then one prime integral is found. And for the second one I tried to multiply the first one by $x$ and the second one by $y$ and sum them to get: $$\frac{xdx+ydy}{2z(x^2+y^2)}$$ Also if I equal this with the third one: $$\frac{d\left(\frac{x^2+y^2}{2}\right)}{2z(x^2+y^2)}=\frac{dz}{z^2-(x^2+y^2)}$$ But I dont know how to finish this one.

Answer 1

The first integral should give

$$ \frac{dx}{x} = \frac{dy}{y} \implies y = c_1 x \implies c_1 = \frac{y}{x} $$

For the second, let $r^2 = x^2+y^2$, then $rdr = xdx + ydy$, so you have

$$ \frac{dr}{2zr} = \frac{dz}{z^2-r^2} $$

or $$ \frac{dz}{dr} = \frac{z^2-r^2}{2zr} = \frac12 \left(\frac{z}{r} - \frac{r}{z}\right) $$

This is homogeneous, so you can substitute $z = ur$ to get

$$ u + r\frac{du}{dr} = \frac12 \left(u - \frac{1}{u}\right) $$

$$ \implies \frac{du}{dr} = -\frac{1+u^2}{2ur} $$

$$ \implies \frac{2udu}{1+u^2} = -\frac{dr}{r} $$

$$ \implies 1+u^2 = \frac{c_2}{r} $$

$$ \implies z^2 = c_2r-r^2 $$

Finally, we set $c_2 = f(c_1)$ to obtain

$$ z(x,y) = \sqrt{\sqrt{x^2+y^2}f\left(\frac{y}{x}\right)-x^2+y^2} $$