<s>[INST] Find the orthogonal trajectories of the family of curves 
Find the orthogonal trajectories of the family of curves $y^5=kx^2$.

I start by knowing that I need a differential equation satisfying all members of family, which means $k$ needs to be eliminated.
Differentiating both sides with respect to $x$:
\begin{align*}
y^5 &= kx^2\\
5y^4 \frac{dy}{dx} &= 2kx\quad\text{Replacing }k\\
5y^4 \frac{dy}{dx} &= 2 \frac{y^5}{x^2}x\\
\frac{5}{y} \frac{dy}{dx} &= \frac{2}{x}\\
\frac{dy}{dx} &= \frac{2}{5} \frac{y}{x}
\end{align*}
Now I have a differential equation, and the orthogonal trajectory should be represented by the negative reciprocal.
\begin{align*}
\frac{dy}{dx}&=-\frac{5}{2}\frac{x}{y}\\
\int ydy&=\int-\frac{5}{2}xdx\\
y&=-\frac{5}{4}x^2+C
\end{align*}
This equation looks orthogonal to me if I plot it, but all the valid answers are of a higher order (e.g., $y^2+\frac{5}{2}x^2=C$). There must be a gap in my understanding somewhere.

https://www.desmos.com/calculator/v6on063jfn
 [/INST] I made a mistake when taking $\int ydy$. It should be $\frac{y^2}{2}$ not $y$.
This means that,
$$
\frac{y^2}{2}=-\frac{5}{4}x^2 + C\\
\frac{y^2}{2}+\frac{5}{4}x^2=C\\
y^2+\frac{5}{2}x^2=2C=C_2
$$
Which is the correct answer.
</s>