Answer to Question #258533 in Differential Equations for Anuj

Question

dx/(z^2+2y)=dy/(z^2+2x)=dz/-z

Answer 1

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

$ln(x-y)/2=lnz+lnc'_1$

$c_1=\frac{x-y}{z^2}$

$\frac{xdx-ydy}{z^2(x-y)}=\frac{xdx-ydy}{c_1z^4}=\frac{dz}{-z}$

$x^2-y^2=-2c_1z^4/4=-c_1z^4/2$

$c_2=-c_1/2=\frac{x^2-y^2}{z^4}$

$F(c_1,c_2)=F(\frac{x-y}{z^2}, \frac{x^2-y^2}{z^4})=0$