TITLE: Solutions to the equation $y^{(n)} y = 1$ for even $n$
QUESTION [8 upvotes]: 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.

REPLY [9 votes]: Let $y=y(x)$ and write primes for derivatives w.r.t $x$. In the equation $$y''y=1$$ the independent variable $x$ does not appear, so we can do a (somewhat magical) change of variables: in the new equation, the independent variable will be $y$ and the dependent variable will be $p=y'$. If we use dots to denote derivatives with respect to $y$, a little computation using the chain rule shows that $p''=p\dot p$, so the equation becomes $$\dot ppy=1.$$ This can be rewritten as $$p\,\mathrm dp=\frac{\mathrm dy}{y},$$ which has its variables separated. It can be integrated at once, to get the relation $$\frac{p^2}{2}=\log y+c$$ with $c$ a constant. Now we recall what $p$ was, that is $y'$, and this becomes a new (first order!) ODE on $y(x)$. We have «simplified» things... (We could have obtained the same reduction od order by multiplying the original equation by $y'$, dividing it by $y$ and integrating: the trick with the $p$ substitution is, though, a general method)
The same approach will similarly simplify the general case.