Partial differential equation - integral surface I have to find the integral surface of the PDE
$$x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$$
containing the straight line $x+y=0, z=1$.
From the auxiliary equations, I can obtain $xyz=C_{1}$ for some constant $C_{1}$. To proceed further, I need to find another relation involving $x,y,z$ or at least one of them. However, I am having difficulty finding this. Would someone be able to help me determine the other relation?
 A: $x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$ is ambiguous because the symbols are not defined. I suppose that the PDE is :
$$x(y^2+z)\frac{\partial z}{\partial x} - y(x^2+z)\frac{\partial z}{\partial y} = (x^2-y^2)z(x,y)\tag 1$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{(x^2-y^2)z}$$
Search for a first characteristic equation :
$$\frac{ydx+xdy}{y(x(y^2+z))+x(-y(x^2+z))}=\frac{dz}{(x^2-y^2)z}=\frac{d(xy)}{xy(y^2-x^2)}$$
$$\frac{dz}{z}=-\frac{d(xy)}{xy}$$
$$xyz=c_1$$
Seach for a second characteristic equation :
$$\frac{xdx+ydy}{x(x(y^2+z))+y(-y(x^2+z))}=\frac{dz}{(x^2-y^2)z}=\frac{d(x^2+y^2)/2}{(x^2-y^2)z}$$
$$dz=d(x^2+y^2)/2$$
$$z-\frac12 (x^2+y^2)=c_2$$
The general solution of the PDE on implicit form $c_2=F(c_1)$ is :
$$z-\frac12 (x^2+y^2)=F(xyz)\tag 2$$
$F$ is an arbitrary function (to be determined according to the boundary condition).
Condition : $z=1$ on $y=-x$
$$1-\frac12 (x^2+(-x)^2)=F(x(-x))$$
$$F(-x^2)=1-x^2$$
Let $X=-x^2$
$$F(X)=1+X$$
Now the function $F$ is determined. We put it into the general solution Eq.$(2)$ where $X=xyz$
$$z-\frac12 (x^2+y^2)=1+xyz$$
$$z(x,y)=\frac{1+\frac12 (x^2+y^2)}{1-xy}$$
