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Is there any solution to the differential equation $f'(x) = f(f(x))$?

I couldn't find any information on this kind of DE

Nada F.
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    $f(x)=0$ is a solution. – Arthur Jul 25 '16 at 20:55
  • There's a solution of the form $f(x)=cx^\alpha$. Can you find $c$ and $\alpha$? – Berci Jul 25 '16 at 21:08
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    @Berci not if $\alpha$ is a real number. – Arthur Jul 25 '16 at 21:21
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    If a nonzero solution exists with a fixed point $f(x^)=x^$ (with $x^\neq 0$) then you can find it with a series solution $f(x) = \sum_{k=0}^{\infty} \frac{(x-x^)^kf^{(k)}(x^)}{k!}$. You can compute $f'(x^)=f(f(x^))=f(x^)=x^$, $f''(x^)=f'(f(x^))f'(x^)=f'(x^)^2 = (x^)^2$, and so on. – Michael Jul 25 '16 at 22:06

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