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We all know that derivative of $e^x$ is $e^x$. Is exponential function only function that has such property? If yes how to prove that there are no other functions. If no, what are other functions? Help me please


marked as duplicate by David Mitra, Lost1, user1337, TMM, Davide Giraudo Jan 20 '14 at 13:56

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You seek to solve the ODE $y'=y$ for arbitrary boundary conditions. This is separable and yields $$1 = \frac y{y'}$$ Integration gives $$x+c = \ln(y(x))$$ or $$y(x)=e^{x+c} = e^c e^x = \tilde c e^x$$ The uniqueness is guaranteed by Picard-Lindelöf.

  • $\begingroup$ Isn't f(x)=0 also its own derivative, even though there is no c such that e^c=0? $\endgroup$ – Jarred Allen Aug 24 '16 at 23:37
  • 2
    $\begingroup$ @JarredAllen Yes but that means $y'=0$ so the separation was already invalid in this special case. Additionally, $y < 0$ needs to be dealt with by premultiplying with $-1$. In the end these cases can be dealt with by allowing $\tilde c\in \mathbb R$ instead of $\tilde c > 0$ in the general solution. $\endgroup$ – AlexR Aug 25 '16 at 8:15

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