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: $\displaystyle \frac{z^2}{2}\pm \left\{\frac{z}{2}\sqrt{z^2-4a^2}-2a^2\log \left(z+\sqrt{z^2-4a^2}\right)\right\}=2ax+2y+b$
On
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.
Let's complete Austin20's solution. Substituting $q=ap$ in $dz=pdx+qdy$, we obtain $$ dz=p(dx+ady)=pd\xi \implies p=\frac{dz}{d\xi}\qquad(\xi:=x+ay). \tag{1} $$ Now, let's substitute $q=ap$ in the PDE $1+p^2=qz$ and solve for $p$: $$ 1+p^2=apz \implies p=\frac{az\pm\sqrt{a^2z^2-4}}{2}. \tag{2} $$ Substituting $(2)$ in $(1)$ and integrating, we obtain $$ \frac{dz}{d\xi}=\frac{az\pm\sqrt{a^2z^2-4}}{2} \implies \xi+b=\int\frac{2\,dz}{az\pm\sqrt{a^2z^2-4}}=\int\frac{az\mp\sqrt{a^2z^2-4}}{2}dz $$ \begin{align} \implies 2\alpha x+2y+\beta&=\int\left(z\mp\sqrt{z^2-4\alpha^2}\right)dz \\ &=\frac{z^2}{2}\mp\left[\frac{z}{2}\sqrt{z^2-4\alpha^2}-2\alpha^2\log\left(z+\sqrt{z^2-4\alpha^2}\right)\right], \tag{3} \end{align} where $\alpha:=\frac{1}{a}$ and $\beta:=\frac{2b}{a}$.