Answer to Question #266764 in Differential Equations for Zuru

Question

Solve: z(x+2y)p-z(2x+y)q=y^2-x^2

Answer 1

Solution;

$z(x+2y)p-z(2x+y)q=y^2-x^2$

The auxiliary equation are;

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

Considering the first two equations;

$\frac{dx-dy}{3z(x+y)}=0$ .....(1)

$dx=dy$

Integrating;

$x=y+c_1$

Considering (1) and the last auxillary equation;

$\frac{dx-dy}{3z(x+y)}=\frac{dz}{x^2-y^2}$

$(x-y)(dx-dy)=3zdz$

$xdx-xdy-ydx+ydy=3zdz$

Integrating;

$\frac{x^2}{2}-2xy+\frac{y^2}{2}=\frac 32z^2+c_2$

$x^2-4xy+y^2=3z^2+c_2$

The solution of the equation is;

$F(x-y,x^2+y^2-3z^2-4xy)=0$