Solving a non-linear Partial differential equation $px^5-4q^3x^2+6x^2z-2=0$ I have to find out the complete integral of :
$$px^5-4q^3x^2+6x^2z-2=0$$
My attempt: Let $$f(x,y,z,p,q)=px^5-4q^3x^2+6x^2z-2$$ 
So, $$f_p=x^5,f_q=-12q^2x^2,f_x=5px^4-8q^3x+12xz,f_y=0,f_z=6x^2$$
Applying Charpit's method:
$$\frac{dx}{x^5}=\frac{dy}{-12q^2x^2}=\frac{dz}{px^5-12q^3x^2}=\frac{-dp}{5px^4-8q^3x+12xz+6px^2}=\frac{dq}{12x^2q^2} $$ $$\implies \frac{dy}{-12q^2x^2}=\frac{dq}{12x^2q^2}$$ $$\implies q=\sqrt{(y+c)}$$

where $c$ is an arbitrary constant.

I don't see here a way to find out $p$. Please help me solve this problem. Even a hint would be much help.

I'll appreciate any help towards it.Thanks in advance.
Hint:

$x^5z_x-4x^2z_y^3+6x^2z-2=0$

$x^3z_x-4z_y^3+6z-\dfrac{2}{x^2}=0$

$x^3z_{xy}-12z_y^2z_{yy}+6z_y=0$

Let $u=z_y$ ,

Then $x^3u_x-12u^2u_y+6u=0$

$12u^2u_y-x^3u_x=6u$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{du}{dt}=6u$ , letting $u(0)=1$ , we have $u=e^{6t}$

$\dfrac{dy}{dt}=12u^2=12e^{12t}$ , we have $y=e^{12t}+y_0=u^2+y_0$

$\dfrac{dx}{dt}=-x^3$ , we have $\dfrac{1}{x^2}=2t+f(y_0)=\dfrac{\ln u}{3}+f(y-u^2)$ , i.e. $u=e^\frac{3}{x^2}F(y-u^2)$ , i.e. $u^2=y-g\left(ue^{-\frac{3}{x^2}}\right)$
From given pde and by using the value q, we get 
$P=\frac{4(y+c)^{\frac{3}{2}}}{x^3}-\frac{6z}{x^3}+\frac{2}{x^5}$
and subtitute the values $p$ and $q$ in $Z=pdx+qdy$ and use multiplier $e^{-\frac{3}{x^2}}$ 
and on integration we get complete required complete integral