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:
$$\frac{dx}{x^5}=\frac{dy}{-12q^2x^2}=\frac{dz}{px^5-12q^3x^2}=\frac{-dp}{5px^4-8q^3x+12xz+6px^2}=\frac{-dq}{6x^2q} $$ $$\implies \frac{dy}{-12q^2x^2}=\frac{dq}{-6x^2q}$$ $$\implies q=\sqrt{(y+c)}$$

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. $u^2=y-g\left(ue^{-\frac{3}{x^2}}\right)$
