$$\frac{d}{dx} ce^x = ce^x$$ 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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asked Mar 11, 2014 at 3:53
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$\begingroup$ Trivial function $f(x)=0$ also works. $\endgroup$– John HabertMar 11, 2014 at 3:55
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5$\begingroup$ That corresponds to $c=0$. $\endgroup$– vadim123Mar 11, 2014 at 3:55
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$\begingroup$ The answer depends on your domain... If the domain is disconected, the answer is no ;) $\endgroup$– N. S.Mar 11, 2014 at 4:15
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$\begingroup$ @N.S. Can you elaborate on this? $\endgroup$– Cameron MartinMar 11, 2014 at 4:17
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$\begingroup$ 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. $\endgroup$– N. S.Mar 11, 2014 at 4:18
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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.
answered Mar 11, 2014 at 4:09