Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer

\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 $D$.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.