7

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

AlexR
  • 24,905
math
  • 73
  • I can't add another answer, but it's also true without having to go to ODE theory: If f(x) is a function equal to its own derivative, then $$\frac{d}{dx}[f(x)e^{-x}] = f'(x)e^{-x} - f(x)e^{-x}= 0$$ Thus $f(x)e^{-x}$ is a constant $C$, or $$f = Ce^{x}$$ – Keshav Sep 28 '23 at 17:49

1 Answers1

13

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.

AlexR
  • 24,905
  • 1
    Isn't f(x)=0 also its own derivative, even though there is no c such that e^c=0? – Jarred Allen Aug 24 '16 at 23:37
  • 3
    @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. – AlexR Aug 25 '16 at 08:15