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.

How to find $f(x)$ if $\frac{df(x)}{dx}=f(x)$? I know $c e^x$ is a solution, but how does one find it and how to prove it is the complete solution?

share|improve this question

3 Answers 3

up vote 5 down vote accepted

To find it, you can use separation of variables. If you're accustomed to recognizing that $\frac{f'}{f}=\log(|f|)'$, you can integrate directly as Chandru1 noted. Otherwise, you can use the substitution $u=f$.

Even without knowing what the solutions are, note that if $f$ and $g$ are solutions and $g$ is nonzero, then $\left(\frac{f}{g}\right)'=\frac{f'g-g'f}{g^2}=0$, so $\frac{f}{g}$ is constant, which means $f=cg$ for some constant $c$. This shows that the multiples of one nonzero solution form the complete solution. In general, a first order homogeneous linear ODE has a one dimensional solution space.

share|improve this answer

Assume $f:\mathbb{R} \to \mathbb{R}$ and the equation if valid for all $x \in \mathbb{R}$.

Trying to divide by $f$ or trying to talk in terms of $\log f$ would need more justification (what if $f$ is zero in some interval?) and has the potential to lose some solutions (though I am pretty sure there will be some theory there which will justify the correctness of the method).

A simpler way, which avoids taking cases:

$$f'(x) = f(x) \iff e^{-x} f'(x) - e^{-x} f(x) = 0 \iff (e^{-x} f(x))' = 0$$

$$e^{-x} f(x) = c$$

share|improve this answer

Just integrate. You have $$\int \frac{f'(x)}{f(x)} \ \textrm{dx} = \textrm{dx}$$ therefore you have $\log{f(x)} = x + C$, So $f(x)=e^{x+C}=e^{x}.e^{C}=ke^{x}$.

share|improve this answer
1  
Ok, but how do you integrate f'(x)/f(x) ? –  user1708 Mar 19 '11 at 5:44
    
@Solomoan: Isn't it $\log{f(x)}$. Differentiate $\log{f(x)}$ what do you get. $\frac{1}{f(x)} \cdot f'(x)$ –  anonymous Mar 19 '11 at 5:46
4  
You left out the possibility that $k<0$. –  Jonas Meyer Mar 19 '11 at 5:53

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.