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?
$x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$ 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 $X=xyz$ $$z-\frac12 (x^2+y^2)=1+xyz$$ $$z(x,y)=\frac{1+\frac12 (x^2+y^2)}{1-xy}$$
$$\frac{dx}{x(y^{2}+z)} = \frac{dy}{-y(x^{2}+z)} = \frac{dz}{z(x^{2}-y^{2})}$$
Multiplying the fraction in $dx$ by $x/x$, the fraction in $dy$ by $y/y$ and using componendo-dividendo on the equality between the ratios in $dx$ and $dy$, we get
$$\frac{dz}{z(x^{2}-y^{2})}=\frac{xdx+ydy}{x^{2}y^{2}+x^{2}z-y^{2}x^{2}-y^{2}z}=\frac{xdx+ydy}{z(x^{2}-y^{2})}$$
and hence we have the next characteristic curve
$$z - \frac{1}{2} \left(x^{2}+y^{2} \right) = C_{2}$$
– Matthew Cassell Dec 07 '21 at 04:28