I recently came across this equation : $\forall x \in \mathbb{R}_+^*, f'(x) = f\left(\frac1{x}\right)$where $f \in \mathcal{C}^1(\mathbb{R}, \mathbb{R})$.
I've done the following, but I'm stuck at the end. Could you give me pointers? Thanks!
Differentiating yields $\forall x, f''(x) = -\frac1{x^2}f(x) \tag{$S_0$}$Solutions in the form $x \mapsto \frac1{x^\phi}$ work iff $\phi(\phi+1) = -1 $, ie. $\phi = \frac{-1 \pm i \sqrt{3}}{2} =e^{\pm 2i\pi/3} = j, \overline{j}$. Elements of the vector space generated by the free pair $(x^j, x^\overline{j})$ are therefore solutions of ($S_0$).
I then feed $\lambda x^j + \mu x^\overline{j}$ in the original equation, which yields $-\lambda j\frac1{x^{j+1}}-\mu\overline{j}\frac1{x^{\overline{j} + 1}} = \frac{x^{j + \overline{j}}}{\lambda x^\overline{j} + \mu x^j}$, then $(-\lambda j x^{\overline{j}+1} - \mu \overline{j} x^{j+1})(\lambda x^{\overline{j}} + \mu x^j) = x^{1+j+\overline{j}} = x^0 = 1$, and $-\lambda^2 j x^{2\overline{j} + 1} - \mu^2 \overline{j} x^{2j+1} - \lambda\mu(j + \overline{j}) = 0 $. Thus, $ \lambda^2 j x^{-2i\sin(2\pi/3)} + \mu^2 \overline{j} x^{2i\sin(2\pi/3)} = \lambda\mu$
Does that mean that no solutions can be found to the original equation, except the trivial $x \mapsto 0$ one? Or that I didn't take the right approach? I can't figure out how to handle the last equality.