A long time ago I was curious about the closed-form solutions to the equation:
\begin{equation*} \frac{d^{n}y}{dx^n} y = 1. \end{equation*}
For $n$ an odd number, try $y = A x^k$. Then $y^{(n)} = A k(k-1)...(k-n)x^{k-n}$. This gives the formula
\begin{equation*} A^2 k(k-1)...(k-n) x^{2k - n - 1} = 1 \end{equation*}
which can only be true if $k = \frac{n+1}{2}$ and $n$ is odd ($k$ cannot be an integer for the formula to work, check this yourself). Furthermore one has to have that
\begin{equation*} A = (k(k-1)...(k-n))^{-1/2} \end{equation*}
which is real if $n$ is odd.
Thus there are closed-form solutions to my problem for $n$ odd, and my question is if anyone can find a closed-form solution for $n=2$ or in general if $n$ is even.