TITLE: Find the complete integral using Charpit Method : $2x(q^2z^2+1) = pz$.
QUESTION [0 upvotes]: Problem : Solve the  integral using Charpit Method  $2x(q^2z^2+1) = pz$ to show that $z^{2}=2\left(a^{2}+1\right) x^{2}+2 a y+b$.
My efforts: 
The given equation is 
$$
F=2x(q^2z^2+1) - pz=0.
$$
Charpit's equation:
Charpit's auxiliary equations are 
\begin{align*}
& \dfrac{dp}{F_x+pF_z}=\dfrac{dq}{F_y+qF_z}=\dfrac{dz}{-pF_p-qF_q}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}\\
\implies &  \frac{dp}{2q^2z^2+2+p\cdot (4xq^2z-p)} = \frac{dq}{0+q(4xq^2z-p)}= \frac{dz}{-p(-z)-q(4xqz^2)} =\frac{dx}{z} = \frac{dy}{-4xqz^2} 
\end{align*}
What can I do now?

REPLY [2 votes]: $$2x\left(\frac{\partial z}{\partial y}\right)^2z^2+2x-\frac{\partial z}{\partial x}z=0 \tag 1$$
It is easy to prove that 
$$z^2=(a^2+1)x^2+a\,y+b \tag 2$$ 
is solution of Eq.$(1)$ :
Differentiating $(2)$ gives $\quad 2z\frac{\partial z}{\partial x}=2(a^2+1)x\quad$ and $\quad 2z\frac{\partial z}{\partial y}=a.\quad$ Thus 
$$\quad 2x\left(\frac{\partial z}{\partial y}\right)^2z^2-\frac{\partial z}{\partial x}z=2x\big(a\big)^2+2x-\big(2(a^2+1)x\big)=0$$
This is not what it is asked in the question.
It is asked " Solve the integral using Charpit Method". But what integral ? 
Moreover the Charpit Method leads to a particular solution if some boundary conditions are specified. Since no boudary condition is given in the wording of the question, why do you expect the solution $(2)$ more specifically than another one ?
In order to help you, think about the simpler PDE below :
Change of variables : $\quad\begin{cases}
X=x^2 \\
Z=z^2
\end{cases}$
The PDE $(1)$ is transformed into PDE $(3)$:
$$\left(\frac{\partial Z}{\partial y}\right)^2+4-2\frac{\partial Z}{\partial X}=0 \tag 3$$
With $P=\frac{\partial Z}{\partial X}$ and $Q=\frac{\partial Z}{\partial y}$
$$Q^2-2P+4=0$$
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