Solving PDEs using Charpit's Method To Solve : $\displaystyle 1+p^2=qz$
I have solved this equation till auxiliary equation:
$\displaystyle \frac{dp}{-pq}=\frac{dq}{-q^2}=\frac{dz}{2p^2-qz}=\frac{dx}{2p}=\frac{dy}{z} $  
p = ∂z/∂x
q = ∂z/∂y

Now, I can't think ahead ..
The given answer is: $\displaystyle \frac{z^2}{2}\pm \left\{\frac{z}{2}\sqrt{z^2-4a^2}-2a^2\log \left(z+\sqrt{z^2-4a^2}\right)\right\}=2ax+2y+b$

Let $f(z,p,q) = 1 + p^2 -qz = 0  ...... (1)$
Then $F_p = 2p$,$F_q = -z$, $F_z = -q$,
Therefore the Charpit's Equations are
$\frac{dx}{2p} =\frac{dy}{-z} =\frac{dz}{2p^2-qz} =\frac{dp}{pq}=\frac{dq}{q^2}$
Then $\frac{dp}{pq} = \frac{dq}{q^2} => ln|q| = ln|p|+ln|a|$, where a is constant
$=>$ $q=ap$
From equation $(1)$  you will find the value of p then using these values and from the equation $dz=pdx +qdy$ 
You'll get the required complete integral.
