I have been solving a pde for real $a$,$b$.

$pq^2=ax+by$, **where** $p=\frac{dz}{dx}$, $q=\frac{dz}{dy}$.

I am required to find the first integral of the equation: $F(z,x,y,c\_1,c\_2)=0$

I wrote down the Charpit's/auxiliary equations:

$\frac{dp}{a}=\frac{dq}{b}=\frac{dz}{3q^2p}=\frac{dx}{q^2}=\frac{dy}{2qp}$.

In the course of solving which I have arrived at **the following system of equations (1)**:

$dz=pdx+qdy$

$z=\frac{3b^2p^4}{4a^3}-\frac{2bc\_1p^3}{a^3}+\frac{3c\_1^2p^2}{2a^3}+c\_5$;

$x=\frac{b^2p^3}{3a^3}-\frac{bc\_1p^2}{a^3}+\frac{(c\_1)^2p}{a^3}+c\_4$;

$y=\frac{2bp^3}{3a^2}-\frac{c\_1p^2}{a^2}-c\_3$,

where $c\_1,c\_3,c\_4$ and $c\_5$ are constants which should add up to two constants $c\_1$ and $c\_2$ in the end. As you can see, this system of equations is only not resolved with respect to $p$, which is the only term I haven't been able to write in the form of $g(z,x,y)$. 

The system (1) is correct when substituted into the equation, I've checked. **The only fraction i haven't used in any way is the third one: $\frac{dz}{3q^2p}=\frac{dz}{3(\frac{bp-c\_1}{a})^2p}$.**

**The equation for $x$ is a non-linear first order ODE, because there is no $y$ in the equation, but Wolfram Alpha does not solve it, for whatever reason. If it can be solved with respect to $z$ ($p=\frac{dz}{dx}$), then my problem is solved. How could i do that?** If it can be done, then my immediate follow-up question is the following: why didn't i need to use the fraction $\frac{dz}{3q^2p}=\frac{dz}{3(\frac{bp-c\_1}{a})^2p}$ in the course of solving the problem?

I had also obtained the following equations:

$bp=aq+c\_1$ or $q=\frac{bp-c\_1}{a}$,

$dy=dp(\frac{2bp^2}{a^2}-\frac{2pc\_1}{a^2})$ and 

$dx=\frac{dp}{a}(\frac{bp-c\_1}{a})^2$.

I've tried to somehow arrive at $z=f(x,y,c\_1,c\_2)$ or $F(z,x,y,c\_1,c\_2)=0$, but had no success because the system (1) is not nicely linear and terms with $p$ always crawl into calculations. I'm now out of ideas.

Could anybody perhaps help me in some way? Or help me solve the ODE $x=\frac{b^2p^3}{3a^3}-\frac{bc\_1p^2}{a^3}+\frac{(c\_1)^2p}{a^3}+c\_4$? Thank you!