General solution of given first order PDE

Question

Find the general solution of given equation: $$ pz - qz = z^2 + (x + y)^2 ,~~p=\frac{∂z}{∂x},~q=\frac{∂z}{∂y} $$ I got one solution $ x + y = c$ , $c$ is a constant. What is the other solution. Please help.

Answer 1

As your equation is linear, you can determine the characteristic curves by solving $$ \dot x=a(x,y,z)=z\\ \dot y=b(x,y,z)=-z\\ \dot z=c(x,y,z)=z^2+(x+y)^2 $$ or $$ \frac{dx}{z}=-\frac{dy}{z}=\frac{dz}{z^2+(x+y)^2} $$ As you found, the first two equations combine to $$x+y=c_1.$$ Combining the first and third equations gives $$\frac{zdz}{z^2+c_1^2}=dx\implies z^2+c_1^2=c_2e^{2x}.$$

With $c_2=\phi(c_1)$ you get the general solution as $$z=\pm\sqrt{\phi(x+y)e^{2x}-(x+y)^2}$$