$$\frac{d}{dx} ce^x = ce^x$$ Are there any other functions $f$ such that $$\frac{d}{dx} f(x) = f(x)$$ or is $ f(x) = ce^x $ the only one?
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Trivial function $f(x)=0$ also works. – John Habert Mar 11 '14 at 03:55
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5That corresponds to $c=0$. – vadim123 Mar 11 '14 at 03:55
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The answer depends on your domain... If the domain is disconected, the answer is no ;) – N. S. Mar 11 '14 at 04:15
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@N.S. Can you elaborate on this? – Cameron Martin Mar 11 '14 at 04:17
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If your domain is $(-\infty,0) \cup (0, \infty)$ for example, on each of these intervals you get a function of your type, but the constants can be different. – N. S. Mar 11 '14 at 04:18
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What do you mean by the constants will be different? Wouldn't you get the function $ce^x$, but defined on the domain $(-\infty, 0) \cup (0, \infty)$ when you differentiate it? – Cameron Martin Mar 11 '14 at 04:41
2 Answers
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No: If $f$ were such a function, consider $g(x) = f(x) e^{-x}$. Then
$$g'(x) = f'(x) e^{-x} + f(x) (-e^{-x}) = f(x) e^{-x} - f(x) e^{-x} = 0$$
As a result, $g$ is constant.
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Consider the first order differential equation $$f'(x)=f(x)$$ which is separable. It's integration leads to $$f(x)=c e^x$$ and this is the only possible solution.
Claude Leibovici
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