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Prove that the only differentiable function $f \colon \mathbb R \to \mathbb R$ such that $f^\prime(x)=f(x) \mspace{1ex} \forall x\in \mathbb R$ is $f(x)=ce^{x}, \forall x\in \mathbb R$, and for some $c\in \mathbb R$.

Let $g:\mathbb R \to \mathbb R$, $g(x)=f(x)/e^{x}$. Then $$g^\prime(x)=\frac{e^{x}f^\prime(x)-f(x)e^{x}}{e^{2x}}=\frac{f^\prime(x)-f(x)}{e^{x}};$$

by hypothesis $f^\prime(x)=f(x)$ hence $g^\prime(x)=0$, and therefore $g(x)=c$ for some $c\in \mathbb R$.

Then $f(x)=ce^{x}, \forall x\in \mathbb R$, and for some $c\in \mathbb R$.

I would like you to tell me if this proof is correct thank you :)

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    $\begingroup$ The proof is good. $\endgroup$ May 29, 2014 at 1:12
  • $\begingroup$ The term is "differentiable", not "derivable". $\endgroup$
    – Batman
    May 29, 2014 at 1:33
  • $\begingroup$ You have some incorrect parens. For example, $g'((x)$ should be $g'(x)$. $\endgroup$ May 29, 2014 at 2:16
  • $\begingroup$ This method of proof is called: "integrating factor". In your case, the integrating factor is $e^{-x}$. $\endgroup$
    – GEdgar
    Sep 8, 2017 at 16:04

2 Answers 2

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The proof you have provided is good.

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$f(x) = 0$

Is also a function with this property.

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    $\begingroup$ This function is of the OP's form, $ce^x$, with $c=0$. $\endgroup$ Sep 8, 2017 at 15:39
  • $\begingroup$ That's counted when $c=0$. $\endgroup$ Sep 8, 2017 at 15:39

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