<s>[INST] Finding the orthogonal trajectories of the family of cissoids Find the orthogonal trajectories of the cissoids, $y^2=\frac{x^3}{a-x}$.
My solution:
$\left( a-x\right)y^2=x^3$
$\left( a-x\right)2y dy-(2x^2+y^2)dx=0$
$\left( a-x\right)2ydx + (2x^2+y^2)dy=0$
$a=\frac{x^3}{y^2}+x$
$2y \left( \frac{x^3}{y^2}\right)dx+(2x^2+y^2)dy=0$
$2x^3dx+(2x^2+y^2)dy=0$
I do not know how to go further. What method could be used to solve this differential equation?
 [/INST] Taking a derivative of your equation
\begin{align}
y&=\sqrt{\frac{x^3}{a-x}},\\
\frac{\mathrm dy}{\mathrm dx}&=\frac{3a-2x}{2}\sqrt{\frac{x}{(a-x)^3}},
\end{align}
and substituting $a=x^3/y^2+x$ gives
\begin{align}
\frac{\mathrm dy}{\mathrm dx}&=\frac{3x^3/y^2+x}{2}\sqrt{\frac{x}{(x^3/y^2)^3}}.
\end{align}
The orthogonal curves have slope that's the negative of the inverse, i.e.,
\begin{align}
\frac{\mathrm dy_{\bot}}{\mathrm dx}=\frac{-2}{3x^3/y_{\bot}^2+x}\sqrt{\frac{(x^3/y_{\bot}^2)^3}{x}}.
\end{align}
Substituting $y_{\bot}=xu$ gives that
\begin{align}
x\frac{\mathrm du}{\mathrm dx}+\frac{2+u(3+u^2)}{3+u^2}=0.
\end{align}
This is separable,
\begin{align}
\int\left( \frac{3u+u^3}{2+u(3u+u^3)}\frac{\mathrm du}{\mathrm dx}+\frac{1}{x}\right)\mathrm dx&=\int 0\mathrm dx\\
\ln(u^2+1)-\frac{1}{2}\ln(u^2+2)+\ln x&=k\\
\frac{u^2+1}{\sqrt{u^2+2}}&=\frac{k}{x}.
\end{align}
Solving for $y_{\bot}=u/x$,
\begin{align}
y_{\bot}^2=k\pm\sqrt{k(2x^2+k)}-x^2.
\end{align}
Fun problem!
</s>