Answer to Question #266343 in Differential Equations for Happy

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 auxiliary equatoin:

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

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

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

integrating:

$\frac{x^2}{2}-2xy+\frac{y^2}{2}=\frac{3}{2}z^2+c_2;$

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

Answer:

The solution of the equation is:

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