<s>[INST] Finding the orthogonal trajectories of $y^2=x^2(1-cx)$ Here is my initial attempt:
$c=\frac{x^2-y^2}{x^3}$
$2y\frac{dy}{dx}=2x-3cx^2$
$-2y\frac{dx}{dy}=2x-\frac{3x^2-3y^2}{x}$
$-2xydx=(2x^2-3x^2-3y^2)dy$
$-2xydx=(-x^2-3y^2)dy$
$2xydx=(x^2+3y^2)dy$
$\frac{dy}{dx}=\frac{2xy}{x^2+3y^2}$
Let $y=ux$; $\frac{dy}{dx}=u+x\frac{du}{dx}$
$u+x\frac{du}{dx}=\frac{2ux^2}{x^2+3u^2x^2}$
$u+x\frac{du}{dx}=\frac{2ux^2}{x^2(1+3u^2)}$
$u+x\frac{du}{dx}=\frac{2u}{1+3u^2}$
$x\frac{du}{dx}=\frac{u-3u^3}{3u^2+1}$
$\frac{dx}{x}=\frac{3u^2+1}{u-3u^3}du$
$\int\frac{dx}{x}=\int\frac{3u^2+1}{u-3u^3}du$
$\ln x=\ln u-\ln (3u^2-1) +\ln K$
$\ln(3u^2x-x)=\ln (uK)$
$3y^2-x^2=yK$
The textbook says that the correct answer is $x^2+3y^2=c_1y$. Can someone find what part of the solution I messed up.
 [/INST] I find always a little sad that such issues aren't associated with a graphical representation in order to have a geometrical feeling of what really happens.
In particular here, I wondered how a family of conical curves (a family of ellipses tangent to the $x$-axis, in red) could have an orthogonal family with third degree curves (with the origin as a double point, in blue and cyan).
This is how.

Please note that the origin is an exceptional point for both families.
This figure has been realized using Matlab but the dynamical way Desmos plots the curves is preferable.
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