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Possible Duplicate:
Proof that $\exp(x)$ is the only function for which $f(x) = f'(x)$

Here's a question I got for homework:

Let f a differentiable function such that $f(x)=f'(x)$ for all $x$. Prove that there exist a $c \in \mathbb{R}$ such that $f(x) = c \cdot e^x$

Hint: notice $\dfrac{f(x)}{e^x}$

So, as it turns out this hint was not enough.

Any more hints? Thanks!

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    What do you get when you differentiate $\dfrac{f(x)}{e^x}$ with respect to $x$?2012-05-22

2 Answers 2

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HINT:

What's the derivative of $\dfrac{f(x)}{e^x}$ (With the quotient rule, perhaps it's easier to see). Then what does that mean?

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It's a first-order linear ordinary differential equation. To put it in a simple way, let $y=f(x)$. Then $ \frac{dy}{dx}=y. $ Hence,
$ \frac{dy}{y}=dx, $ (if $y$ doesn't equal $0$).

Integrating in both sides, we get $ \ln|y|=x+c_1, $ where $c_1$ is a constant. Therefore, $ |y|=e^{x+c_1}=e^{c_1}*e^x. $ Let $|c|=e^{c_1}$, then we get $ y=c*e^x. $ If $y=0$, then it of course satisfies the condition.

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    @ Shane O Rourke The derivative of ln|y| is 1/y, so when you look back to the answer again which I have edited, you will find$c$can be negative. And it can't be the case that$y$happens to take the value zero at some points but not the zero function.(here y is continuous at every x)2012-05-22