Im solvin this equation using charpit method. $$F = z+px+qy - p^2y/2$$
$ dy/y = (-dp)/(2p) $ gives me $p = A/y^2$ and $ (-dp)/(2p) = (-dq)/(2q-p^2/2)$ gives me $dq/dp - (q/p) = - p/4$. On solving I get $$q = -p^2/4 + Bp$$. If I substitute these p and q to get z, I get wrong solution:
$z = -px-qy + p^2y/2$ (from original pde), I get $z=\frac{-Ax}{y^2} + \frac{-BA}{y} + \frac{3A^2}{4y^3}$ which is not correct. (On calculating $z_x = p $ and $z_y = q$ does not satisfy.) Why?
Although $q=-\frac{p^2}{4}+Bp$ is the correct solution to $$ \frac{dp}{2p}=\frac{dq}{2q-\frac{p^2}{2}}, \tag{1} $$ it is not compatible with the integrability condition $p_y=q_x$ if $p=\frac{A}{y^2}$ (don't ask me why). Let me show, however, how to solve the PDE.
Integrating $z_x=p=\frac{A}{y^2}$, we find $$ z=\frac{Ax}{y^2}+f(y). \tag{2} $$ To determine $f(y)$, we substitute $(2)$ in the PDE and use the fact that $q=z_y=-\frac{2Ax}{y^3}+f'(y)$: $$ \left(\frac{Ax}{y^2}+f(y)\right)+\frac{A}{y^2}x+\left(-\frac{2Ax}{y^3}+f'(y)\right)y-\left(\frac{A}{y^2}\right)^2\frac{y}{2}=0 $$ $$ \implies yf'(y)+f(y)=\frac{A^2}{2y^3}. \tag{3} $$ One can easily check that the general solution to $(3)$ is $f(y)=\frac{C}{y}-\frac{A^2}{4y^3}$. Substituting this result in $(2)$, we finally obtain $$ z=\frac{Ax}{y^2}-\frac{A^2}{4y^3}+\frac{C}{y}, \tag{4} $$ where $A$ and $C$ are arbitrary constants.
