Find the complete, general and singular solution of the PDE
\left(\frac{p}{x}\right)^2+\left(\frac{q}{y}\right)^2=\frac{2}{z^2} .
Moreover, find the surface which passes through the line y = 0, z = x – 1.
Firstly, we shall introduce a new unknown function u = u(x, y) by u = z². Then we have
\frac{1}{x^2}\left(\frac{\partial u}{\partial x}\right)^2+\frac{1}{y^2}\left(\frac{\partial u}{\partial y}\right)^2=8
and the complete solution is
u=z^2=a x^2+y^2+\sqrt{2-a^2}+b .
There is no singular solution.
The general solution is
z^2=a x^2+y^2 \sqrt{2-a^2}+f(a), \quad x^2=\frac{a y^2}{\sqrt{2-a^2}}+f^{\prime}(a) .
where f is an arbitrary function of the class C^1 (R).
Let us find the sought after surface. By assumption we have
(x-1)^2=a x^2+f(a),
or
2(x-1)=2 a x, \quad \text { or } \quad x=\frac{1}{1-a} \text {. }
In view of the equation f^{\prime}(a)+x^2=0, we obtain
f(a)=-\frac{1}{a-1} .
Finally, the parametric equation of the Cauchy integral is
z^2=a^2+y^2 \sqrt{2-a^2}-\frac{1}{a-1}, \quad x^2=\frac{a y^2}{2-a^2}-\frac{1}{(a-1)^2} .