3
$\begingroup$

This question already has an answer here:

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

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.