# How to find $f(x)$ if $df(x)/dx=f(x)$

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?

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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.

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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:

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

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

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

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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
You left out the possibility that $k<0$. – Jonas Meyer Mar 19 '11 at 5:53