TITLE: Differential equation related to Golomb's sequence
QUESTION [6 upvotes]: While studying Golomb's sequence, the following differential equation arouse:
$$
f(f(x))=\frac{1}{f'(x)}
$$
I don't know much about differential equations so I am a bit clueless. Is there a way to solve it nicely?
Edit:
With the link provided by Michael Galuza, I could find a function which satisfies the given equation, namely:
$$
f(x)=\varphi^{2-\varphi}x^{\varphi-1}
$$
Where $\varphi=\frac{1+\sqrt5}{2}$ is the golden ratio. However, I did not succeed in proving that this is the only one. Is this feasible?

REPLY [2 votes]: Let $y =f(x)$ then $\frac{dy}{dx}=f'(x)$ so that the D.E. becomes $f(y)=\frac{dx}{dy}$ so $f(y)dy=dx$.
We have two ways:

The D.E. is separable, integrating both sides we get $\int {f(y)dy=x+c}$
The equation $dx-f(y)dy=0$ is exact since $M(x,y)=1$ and $N(x,y)=-f(y)$ we have $M_y=0$ and $N_x=0$.

Let $\Phi(x,y)$ be a differentiable function such that $\Phi_x=1$ $ \Rightarrow $ $\Phi(x,y)=x+h(y)$. On the other hand, $\Phi_y=N(x,y)$ so that $h'(y) = -f(y)$ which gives that $h(y)=-\int{f(y)dy}+C$.
Hence, $\Phi(x,y)=x-\int{f(y)dy}+C$.