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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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marked as duplicate by mixedmath, Michael Greinecker, lhf, The Chaz 2.0, Asaf Karagila May 22 '12 at 12:05

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What do you get when you differentiate $\dfrac{f(x)}{e^x}$ with respect to $x$? –  Brian M. Scott May 22 '12 at 8:27
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2 Answers 2

up vote 5 down vote accepted

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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Sorry for the bad notation. Just new here, and I am still learning how to use the mathematical notation in this website. –  Pan Yan May 22 '12 at 8:42
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I found a useful way to see how people generate symbols here is to click edit and see how they write thing (while not editing the answer of course!) –  Holdsworth88 May 22 '12 at 8:44
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@Holdsworth88: If you right-click on a formula, select Show Math As, and then select TeX Commands, you’ll get a pop-up window showing the $\LaTeX$ that was actually used. –  Brian M. Scott May 22 '12 at 9:00
    
@BrianM.Scott Well, that simplifies that process a bit. Thank you for telling me about that feature. –  Holdsworth88 May 22 '12 at 9:04
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What if $y$ is not the zero function but happens to take the value zero at some points? Also, the constant $c$ needn't be positive, though the argument seems to imply that it is. –  Shane O Rourke May 22 '12 at 10:48
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