$$\frac{d}{dx} ce^x = ce^x$$ Are there any other functions $f$ such that $$\frac{d}{dx} f(x) = f(x)$$ or is $ f(x) = ce^x $ the only one?

  • $\begingroup$ Trivial function $f(x)=0$ also works. $\endgroup$ – John Habert Mar 11 '14 at 3:55
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    $\begingroup$ That corresponds to $c=0$. $\endgroup$ – vadim123 Mar 11 '14 at 3:55
  • $\begingroup$ The answer depends on your domain... If the domain is disconected, the answer is no ;) $\endgroup$ – N. S. Mar 11 '14 at 4:15
  • $\begingroup$ @N.S. Can you elaborate on this? $\endgroup$ – Cameron Martin Mar 11 '14 at 4:17
  • $\begingroup$ 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 '14 at 4:18

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.


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.


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