Find $f:\mathbb R\to\mathbb R$ such that $f(x)=f'(x)$. 
Find $f:\mathbb R\to\mathbb R$ such that $f(x)=f'(x)$.

I know that $f(x)$ can be $e^x$, but how to solve this mathematically? Are there any other solutions? Thank you.
 A: By integrating $f'/f$, you can get to the solution $f(x) = Ce^x$. Now if you want to prove that it is the only solution, my suggestion is as follows:
Hint: Let $f$ be such a function, and let $g(x) = f(x)e^{-x}$. Now calculate $g'(x)$. Can you now prove the result?
A: You have the problem
$$f'(t) = f(t)$$
$$f(0) = x_0$$
The solution you already know is 
$$f(t) = x_0e^t$$
Suppose $Y(t)$ is another solution of the same problem. Consider
$$y(t,s) = e^{(t-s)}Y(s).$$
The map $s \to y(t,s)$ is continuosly differentiable and
$$\frac{\partial y(t,s)}{\partial s} = e^{(t-s)}\frac{dY(s)}{ds} -e^{(t-s)}Y(s) = e^{t-s}Y(s) - e^{t-s}Y(s) = 0.$$
This implis that $y(t,s)$ is constant in $s$ and therefore 
$$x_0e^t = y(t,0) = y(t,t) = Y(t)$$
I hope this can help you!
A: Simple elementary differential equation. Put $\,y=f(x)\,$ , then::
$$f'(x)=f(x)\iff \frac{dy}{dx}=y\implies \int\frac{dy}{y}=\int dx\implies$$
$$\implies \log y=x+c\implies y=e^{x+c}=Ke^x\;,\;\;K:=e^c=\text{ a constant}$$
A: Let $f : \mathbb{R} \to \mathbb{R}$ such that $f'(x)=f(x)$ for all $x \in \mathbb{R}$. Set $g(x)=f(x)e^{-x}$. Then $g'(x)=0$, hence $f(x)=C \cdot e^{x}$ for some $C \in \mathbb{R}$.
