Question

dx/2xz=dy/2yz=dz/z^2-x^2-y^2

Answer 1

$\frac{dx}{2xz}=\frac{dy}{2yz}=\frac{dz}{z^2-x^2-y^2}$

The first equation gives

$\frac{dx}{x}=\frac{dy}{y}$, i.e. $\frac{xdy-ydx}{xy}=\frac{x}{y}d(\frac{y}{x})=0$,

$d(\frac{y}{x})=0$

$\frac{y}{x}=k={\rm const}$

The second equation gives

$\frac{dx}{x}=\frac{2zdz}{z^2-x^2-y^2}=\frac{dz^2}{z^2-(1+k^2)x^2}$

$xdz^2-(z^2-(1+k^2)x^2)dx=0$

$\frac{xdz^2-z^2dx}{x^2}-(1+k^2)dx=0$

$d(\frac{z^2}{x}-(1+k^2)x)=0$

$\frac{z^2}{x}-(1+k^2)x=C$

$z^2-(1+k^2)x^2=Cx$

$z^2-x^2-y^2=Cx$

The general integral surface of the given ODE has an equation determined by an arbitrary differentiable function $\Phi$ of $(k,C)$ :

$\Phi(k,C)=\Phi(\frac{y}{x},\frac{z^2-x^2-y^2}{x})=0$

Answer. $\Phi(\frac{y}{x},\frac{z^2-x^2-y^2}{x})=0$, where $\Phi$ is an arbitrary differentiable function.