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.
 A: Hint:
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dz}{dt}=z$ , letting $z(0)=1$ , we have $z=e^t$
$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0z$
$\dfrac{dx}{dt}=\dfrac{x(z-2y^2)}{z-y^2-2x^3}=\dfrac{x(e^t-2y_0^2e^{2t})}{e^t-y_0^2e^{2t}-2x^3}$
A: From 2nd and third ratio, y=k.z
Put this in first and 2nd ratio and solve. 
The equation is almost in exact form, apart from a factor of -1.
So choose a suitable integrating factor to manage it. 
A: Your calculus is correct.
A first family of characteristic curves comes from $\frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}$ which solution is $z=\frac{1}{c_1}y=c'_1y$
$$\frac{z}{y}=c'_1$$
A second family of characteristic curves comes from 
$$\frac{dx}{x(c'_1y-2y^2)}=\frac{dy}{y(c'_1y-y^2-2x^3)}$$
The solution of this ODE is :
$y=\frac{c'_1}{2}\pm \sqrt{x^3+\frac{(c'_1)^2}{4}+c_2}$
$\left(y-\frac{c'_1}{2}\right)^2-x^3-\frac{(c'_1)^2}{4}=c_2$
$y^2-c'_1y -x^3=c_2$
$$y^2-z-x^3=c_2$$
The general solution of the PDE expressed on the form of implicit equation is :
$$\Phi\left(\:\left(\frac{z}{y}\right)\:,\:\left(y^2-z-x^3\right)\: \right)=0$$
where $\Phi$ is an arbitrary function of two variables.
Or equivalently :
$$z=y\:F(y^2-z-x^3)$$
where $F$ is an arbitrary function.
$\Phi$ or $F$ has to be determined to fit some boundary condition (not specified in the wording of the question).
A: 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 =$$\phi(\frac{y}{z},\frac{z-y^2-2x^2}{x})=0$$
