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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?

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$$2x\left(\frac{\partial z}{\partial y}\right)^2z^2+2x-\frac{\partial z}{\partial x}z=0 \tag 1$$ 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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  • $\begingroup$ @ JJacquelin How did you get equation 1? $\endgroup$
    – BSFU
    May 16 '20 at 4:55
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    $\begingroup$ I didn't get equation (1) : You gave it on the form $2x(q^2z^2+1) = pz$ since $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$. Isn't it ? $\endgroup$
    – JJacquelin
    May 16 '20 at 7:14
  • $\begingroup$ @ JJacquelin I got it now. Thank you very much. $\endgroup$
    – BSFU
    May 16 '20 at 9:39

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