Solve the partial differential equation using lagrange equations

Question

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

I am able to solve the first part and the answer is $x^2-y^2-z^2=u $ The next part I am not able to solve but the ans is $2xy-z^2$

Answer 1

For a first constant of integration we have: $$\dfrac {dx}{x+y}=\dfrac {dy}{x-y}$$ $$\dfrac {d(x-y)}{2y}=\dfrac {dy}{x-y}$$ $$(x-y)^2=2y^2+c_1$$ $$ \implies c_1=x^2-2xy-y^2$$ For the second constant of integration: $$\dfrac {dx}{x+y}=\dfrac {dy}{x-y}=\dfrac {zdz}{x^2+y^2}$$ $$\dfrac {d(x^2-y^2)}{x^2+y^2}=\dfrac {2zdz}{x^2+y^2}$$ $$d(x^2-y^2)=2zdz$$ Integration gives us: $$x^2-y^2=z^2+c_2$$ $$ \implies c_2=x^2-y^2-z^2$$ So the solution is: $$c_2=f(c_1)$$ $$\boxed {z^2=x^2-y^2+f(x^2-2xy-y^2)}$$