Answer to Question #205178 in Differential Equations for Tonny

Question

Solve the following equation.

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

Answer 1

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

$\frac{xdx}{x(y-xz)}=\frac{ydy}{y(yz+x)}=\frac{dz}{x^2+y^2}$

$\frac{xdx-ydy}{-z(x^2+y^2)}=\frac{dz}{x^2+y^2}$

$x^2-y^2+z^2=c_1$

$\frac{ydx+xdy}{x^2+y^2}=\frac{dz}{x^2+y^2}$

$xy-z=c_2$

General solution:

$F(x^2-y^2+z^2,xy-z)=0$