Show by solving $2z+p^2+qy+2y^2=0$ using Charpit's method that $y^2[(x-a)^2+y^2+2z]=b$.
My efforts :
The given equation is
\begin{equation}
F=2z+p^2+qy+2y^2=0.
\end{equation}
Charpit's equation:
Charpit's auxiliary equations are
$ \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}$. This implies
$ \dfrac{dp}{2p}=\dfrac{dq}{3q+4y}=\dfrac{dz}{-2p^2-qy}=\dfrac{dx}{-2p}=\dfrac{dy}{-y}$
Taking 1 st and 4 th fraction we get,
$
\Longrightarrow \frac{d p}{2 p}=\frac{d x}{-2 p}
$
So that $, p=-x+a$.
Now taking 2 nd and 5 th fraction ,we get
\begin{align*}
& \frac{d q}{3 q+4 y}=\frac{d y}{-y} \\
\implies & \frac{d q}{d y}=-\frac{3 q+4 y}{y} \\
\implies & \frac{d q}{d y}+\frac{3}{y} q=-4
\end{align*}
Which is a linear equation in first order.
$
\therefore I . F=e^{\int \frac{3}{y} d y}=e^{3 \log y}=e^{\log y^{3}}=y^{3}
$.
Now, $ q y^{3}=\int(-4) y^{3} d y+b=-y^{4}+b$, where $b$ is a constant. This implies
$ q=-y+b/y^3 $.
Now putting the value of $p$ and $q$ in.
$\begin{array}{l}
d z=p d x+q d y \text { we get } d z=-x(a-x)dx+(b/y^3-y) dy \\
\Longrightarrow z=-(a-x)^2/2-b/(2y^2)-y^2/2+d
\end{array}
$,
where $d$ is a integration constant.
This solution is different from the solution what was given. How can I prove the result?
Best Answer
$$2z+p^2+qy+2y^2=0 \tag 1$$ You found :
$$z=-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2+d \tag 2$$ In order to check your result, put your result (2) into equation (1) :
$p=a-x$
$q=\frac{b}{y^3}$
$$2\left(-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2+d\right)+(a-x)^2+\frac{b}{y^3}y+2y^2=2d$$ This is not $=0$. So your result is false in general (except if $d=0$ ). Thus $$z=-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2 \tag 3$$ $2z=-(a-x)^2-\frac{b}{y^2}-y^2$
$2z+(a-x)^2+y^2=\frac{-b}{y^2}$
$y^2[(x-a)^2+y^2+2z]=-b$
$b$ is an arbitrary constant. You can change it into any other constant. Thus you can change $-b$ into $b$ : $$y^2[(x-a)^2+y^2+2z]=b \tag 4$$ Your solution (2) with $d=0$ is the expected solution.