Partial Differential Equation xp(1+q) = (y+z)q

$p=\dfrac{dz}{dx}$ and $q=\dfrac{dz}{dy}$

I have to find the complete integral solution of the Partial Differential Equation
$$ \Rightarrow f(x,y,z,p,q) = xp + xpq – yq – zq = 0$$
I am trying to solve it using Charpit equations :

For the first and the last fraction: $f_p=x(1+q)$ and $f_y+pf_z=-q(1+q)$; which gives- $$\dfrac{dx}{-x(1+q)}=\dfrac{dq}{-q(1+q)}\Rightarrow \log(x) + \log(a) = \log(q)\Rightarrow q=ax.$$

From the given PDE and $q=ax$, we have $p=\dfrac{a(y+z)}{1+ax}.$

Now for the solution, we have to solve $dz=pdx+qdy \Rightarrow dz=\dfrac{a(y+z)}{1+ax}dx + axdy.$

I am not able to integrate $dz=\dfrac{a(y+z)}{1+ax}dx + axdy$. Any ideas on how to integrate this to get expression for $z$ in terms of $x$ and $y$?

Solutions Collecting From Web of "Partial Differential Equation xp(1+q) = (y+z)q"

Supposing that $q=ax$ and $p=\frac{a(y+z)}{1+ax}$ (I didn’t check the simajinid’calculus).
$$q=\frac{\partial z}{\partial y}=ax\quad\to\quad z=axy+\phi(x)$$

$$p=\frac{\partial z}{\partial x}=ay+\phi'(x)=\frac{a(y+axy+\phi(x))}{1+ax}$$
Of couse, this is only one particular solution, not the general solution of the PDE.