# 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.

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Not quite true, $ke^x$ works for any $k$, but nothing else does. There have been a number of proofs of this posted on MSE. – André Nicolas Jan 29 '13 at 23:22
ok , i will change the question to contain this condition ! @AndréNicolas – Maths Lover Jan 29 '13 at 23:23
@AndréNicolas , can you give me the link of one of these proof ?? give me the simplest one plz :) – Maths Lover Jan 29 '13 at 23:41
The one you accepted is the standard one. – André Nicolas Jan 30 '13 at 1:28

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.

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\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 $D$.

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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 $$f(x) dx = g(y) dy \Rightarrow \int f(x)\,dx = C + \int g(y)\,dy.$$

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