TITLE: Characterization of solutions to $f' = f(1-f)$
QUESTION [7 upvotes]: The sigmoid function $f(x) = \frac{1}{1+e^{-x}}$ has the property that $$f'(x) = f(x)(1-f(x))~~~ and ~~~f(0) = \frac 12$$ 

My question: is $f$ the unique function from $\mathbb R$ to $(0,1)$, perhaps up to some kind of scaling, that satisfies $f' = f(1-f)$?
I don't have much experience with differential equations so a nonlinear one like this is beyond anything I've done before. 
In case it helps, my motivation for this is that this property makes the log likelihood a lot easier in a logistic regression, and I'm wondering if assuming that the inverse link function satisfies this property is equivalent to just taking it to be $f$.

REPLY [4 votes]: Note that since $f(0)=1/2$ then, $f\not\equiv 1$ and , $f\not\equiv 0$ hence a solution cannot an  equilibrium $$f'=f(1-f)\Longleftrightarrow\frac{f'}{f} +\frac{f'}{1-f} =1\Longleftrightarrow \ln (|f|)- \ln(|1-f|) =x+c$$

But,
 $f(x)\in (0,1)$ we obtain 
$$ \ln (f)- \ln(1-f) =x+c\Longleftrightarrow \frac{f}{1-f} =ke^x \Longleftrightarrow  f(x) =1-\frac{1}{1+ ke^x}$$
I am sure you can the $k$ by yourself.