[Math] Solve $z=px+qy+p^2+q^2$ by Charpit’s Method

Question

ind the complete integral of partial differential equation $z=px+qy+p^2+q^2$.
My attempt:
Let $F=px+qy+p^2+q^2=0$. Then by
Charpit's auxiliary equations
$\dfrac{dp}{F_x+pF_z}=\dfrac{dq}{F_y+qF_z}=\dfrac{dz}{-pF_p-qF_q}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}$
we have
$\dfrac{dp}{0}=\dfrac{dq}{0}=\dfrac{dz}{-p(x+2p)-q(y+2q)}=\dfrac{dx}{-(x+2p)}=\dfrac{dy}{-(y+2q)}$ .
This implies $p=a$=constant. Putting this in given equation, $q=\dfrac{-y\pm \sqrt{y^2-4(a^2+ax-z)}}{2}$. What can I do after this to solve the equation?

Answer 1

The given equation is \begin{equation} F=z-px-qy-p^2-q^2=0. \end{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 & \dfrac{dp}{-p+p(1)}=\dfrac{dq}{-q+q(1)}=\dfrac{dz}{-p(-x-2p)-q(-y-2q)}=\dfrac{dx}{-x-2p}=\dfrac{dy}{-y-2q}\\ \implies & \dfrac{dp}{0}=\dfrac{dq}{0}=\dfrac{dz}{px+qy+2p^2+2q^2}=\dfrac{dx}{-x-2p}=\dfrac{dy}{-y-2q} \end{align*} First fraction implies, $dp=0\implies p=a$.
Similarly, let $q=b.$
Now, $dz=pdx+qdy\implies dz=adx+bdy\implies z=ax+by+c$.
Putting the value of $z$ in the given equation,
$ax+by+c=ax+by+a^2+b^2\implies c=a^2+b^2$
Therefore, $z=ax+by+a^2+b^2$.
This is the complete solution.