# Proving that $f(x)=c \cdot e^x$ is the only function such that $f'(x)=f(x)$ [duplicate]

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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## marked as duplicate by mixedmath♦, Michael Greinecker♦, lhf, The Chaz 2.0, Asaf KaragilaMay 22 '12 at 12:05

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

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
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
@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
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