find the general integrals of the given P.D.E

I tried to find the general integrals of the given P.D.E in the yellow box.

And I found $c_1$. But I cannot find another one, say $c_2$.

Please help me finding $c_2$. Thank you.

$$\color{#F3D159}{\boxed{\displaystyle\,\,\color{black}{px(z-2y^2)=(z-9y)(z-y^2-2x^3)}\,\,}} \\ \begin{gather*} P=x(x-2y^2)\\ Q=+y(z-y^2-2x^3)\\ R=z(z-y^2-2x^3) \end{gather*}\\\frac{\mathrm dx}{x(z-2y^2)}=\frac{\mathrm dy}{+y(z-y^2-2x^3)}=\frac{\mathrm dz}{z(z-y^2-2x^3)}\\\frac{\mathrm dy}y=\frac{\mathrm dz}z\implies\ln y=\ln z+\ln c_1\\\boxed{ \frac y z = c_1}$$

• Does this question come from the PDE $x(z-2y^2)\dfrac{\partial z}{\partial x}+y(z-y^2-2x^3)\dfrac{\partial z}{\partial y}=z(z-y^2-2x^3)$ ? – doraemonpaul Nov 2 '13 at 1:12
• Yes, of course!. I forgot writing the expression of $p$ and $q$. Sorry. Can you help me? Dear @Doraemonpaul – user315 Nov 2 '13 at 8:48

$x(z−2y^2)\dfrac{\partial z}{\partial x}+y(z−y^2−2x^3)\dfrac{\partial z}{\partial y}=z(z−y^2−2x^3)$
$\dfrac{x(z−2y^2)}{z−y^2−2x^3}\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}=z$
$\dfrac{dy}{dt}=y$ , letting $y(0)=1$ , we have $y=e^t$
$\dfrac{dz}{dt}=z$ , letting $z(0)=z_0$ , we have $z=z_0e^t=z_0y$
The remaining problem is how to solve $\dfrac{dx}{dt}=\dfrac{x(z−2y^2)}{z−y^2−2x^3}=\dfrac{x(z_0e^t−2e^{2t})}{z_0e^t−e^{2t}−2x^3}$ .