Answer to Question #206651 in Differential Equations for Tanjeel

Question

z(x+y)p+z(x-y)q=x^2+y^2 

Answer 1

$\text{By lagrange method,}\\ \frac{dx}{z(x+y)}=\frac{dy}{z(x-y)}=\frac{dz}{x^2-y^2}\\ first,\\ \frac{dx}{z(x+y)}=\frac{dy}{z(x-y)}\\ \frac{dx}{x+y}=\frac{dy}{x-y}\\ xdx-ydy-xdy-ydx=0\\ xdx-ydy-d(xy)=0\\ \text{Integrating, we get}\\ x^2-y^2-2xy=c_1\\ Second,\\ \frac{xdx-ydy-zdz}{z(x+y)-z(x-y)-z(x^2+y^2)}=0\\ xdx-ydy-zdz=0\\ x^2-y^2-z^2=c_2\\ Therefore,\\ \text{solution is given by,}\\ F(x^2-y^2-2xy,x^2-y^2-z^2)=0$