Find the general integral of $ px(z-2y^2)=(z-qy)(z-y^2-2x^3).$ $ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
Find the general integral of the linear PDE $ px(z-2y^2)=(z-qy)(z-y^2-2x^3). $
My attempt to solve this is as follows:
$ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
$$px(z-2y^2)+qy(z-y^2-2x^3)=z(z-y^2-2x^3)$$
\begin{align*}
\text{The Lagrange's auxiliary equation is:} \frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}=\frac{dz}{z(z-y^2-2x^3)}
\end{align*}
Now consider the 2nd and 3rd ratios, 
\begin{align*}
\frac{dy}{y(z-y^2-2x^3)} & =\frac{dz}{z(z-y^2-2x^3)}\\
\implies \frac{dy}{y} & =\frac{dz}{z}\\
\implies \ln(y) & =\ln(z)+\ln(c_1)\\
\implies \frac{y}{z} & =c_1.
\end{align*}
But I am unable to get the 2nd integral surface. Kindly, help me.
Thanks in advance.

i think you made error in question - in place of $2x^3$ it should be $2x^2$
$$ \frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^2)}=\frac{dz}{z(z-y^2-2x^2)}\Rightarrow1)$$
then take $0,-2y,1$ as multipliers
$$ \frac{-2ydy+dz}{-2y^2(z-y^2-2x^2)+z(z-y^2-2x^2)}=\frac{d(z-y^2)}{(z-2y^2)(z-y^2-2x^2)}\Rightarrow2)$$
Combining fraction $\Rightarrow2)$ with $1 st $ fraction of $\Rightarrow 1) $
$$\frac{dx}{x(z-2y^2)}=\frac{d(z-y^2)}{(z-2y^2)(z-y^2-2x^2)}$$
$$\frac{d(z-y^2)}{dx}=\frac{z-y^2-2x^2}{x} $$
take $z-y^2=u$ we get $$\frac{du}{dx}=\frac{u-2x^2}{x} $$
$$ \frac{du}{dx}-\frac{u}{x}=-2x $$
IF $$ e^{\int\frac{-dx}{x}}=\frac{1}{x} $$
its solution =$$\frac{z-y^2}{x}=-2x+c_{1}$$ or
$$ \frac{z-y^2+2x^2}{x}=c_1$$
final solution =$$\phi(\frac{y}{z},\frac{z-y^2-2x^2}{x})=0$$
