prove: if $y=\frac{dy}{dx}$ then , $y=ce^x$ for some constant $c$ we all know that :
if $y=c e^x$ 
then  $ y= \frac{dy}{dx}$
let $y=f(x)$
now , we want to prove the other way, I mean :
prove,if $y=\frac{dy}{dx}$
then , 
$y=ce^x$ for some constant $c$
can any one prove this? 
I didn't study diffrential equations yet. 
Note, this is not a homework, it's just a question which I want to know its answer :)
so, I don't know if this statement is true or not, but I think that it's true, so I look for its proof which I think will be interesting! Won't it ?
thanks. 
 A: Suppose $y = \frac{dy}{dx}$. Then $$ (y e^{-x})' = \frac{dy}{dx} e^{-x} - e^{-x} y = 0$$
So $ye^{-x}$ is a constant, as desired. 
A: \begin{align}
y&={dy\over dx}\\
dx&={dy\over y}\\
\int dx &=\int {dy\over y}\\
x+C&=\ln|y|\\
e^{x+C}&=e^{\ln|y|}\\
e^x\,e^C&=|y|\\
y&=\pm e^C e^x\\
y&=De^x, \quad D\not=0
\end{align}
but then by inspection $y=0$ is also a solution, so in the end we say $y=De^x$ for any $D$.
A: The general technique is called the "separation of variables." Suppose $y>0$. The idea is to first divide both sides by $y$ to get
$$1 = \frac{1}{y(x)}\frac{dy}{dx}(x).$$
Now integrate both sides:
$$\int_0^x 1\,da = \int_0^x \frac{1}{y(a)} \frac{dy}{dx}(a)\,da.$$
On the right, use the substitution $u = y(a),\ du = \frac{dy}{dx} da$ to get
$$ x = \int_{y(0)}^{y(x)} \frac{1}{u} du  = \log[y(x)] - \log[y(0)]$$
and so
$$y(x) = y(0)e^x.$$
The same trick works whenever you have a differential equation that can be written in the form
$$f(x) = g(y) \frac{dy}{dx}.$$
A notational shortcut that is sometimes used is to "move the $dx$ to the other side" to get
$$f(x) dx = g(y) dy \Rightarrow \int f(x)\,dx = C + \int g(y)\,dy.$$
