Answer to Question #127508 in Differential Equations for Anithra

Question

How to find integral curve of , dx/(xz-y)=dy/(yz-x)=dz/(1-z^2)

Answer 1

$\dfrac{-x*dx+y*dy}{-x*(xz-y)+y*(yz-x)}=\dfrac{dz}{1-z^2}\newline \dfrac{\dfrac{1}{2}d(-x^2+y^2)}{-x^2*z+y^2*z}=\dfrac{dz}{1-z^2}\newline \dfrac{1}{2}\dfrac{d(-x^2+y^2)}{-x^2+y^2}=\dfrac{z*dz}{1-z^2}\newline \dfrac{1}{2}\ln(-x^2+y^2)=-\dfrac{1}{2}\ln \dfrac{(1-z^2)}{C_1}\newline y^2-x^2=\dfrac{C_1}{1-z^2}$