The following equation is given:
\begin{align} f'(x)=f(x)f'(f(x)),x>0 \end{align}
Is there an easy way to solve this equation for $f(x)$?
Thank you
Solving $f'(x)=f(x)f'(f(x)),x>0$ for $f(x)$
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ordinary-differential-equations
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0Well, you could guess-and-check. For example, $f(x) = \mbox{const}$ works ... – 2017-01-10
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0I hope there is a more systematic way to solve this equation. – 2017-01-10
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0$f(x)=\log(x)+c$ seems to be the solution. I tested a few functions ($\exp$, $\log$, ...) and saw that $\log$ works. But hopefully there is another clean way to solve the equation. – 2017-01-11
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0Are you sure? If $f(x) = \log(x) + c$ then $f'(x) = 1/x$ so we have $1/x = (\log(x) + c)\frac{1}{\log(x) + c} = 1$ (when $x\neq e^{-x}$) which is not true for all $x>0$. – 2017-01-11
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0I think $f'(f(x))$ is equal to $\frac{\frac{1}{x}}{\log(x)+c}$, then it works. – 2017-01-11
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1This is not the case: as $f'(x) = 1/x$, we have $f'(f(x)) = 1/f(x) = 1/(\log(x) + c)$. – 2017-01-11
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0You are right:) – 2017-01-11