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The logistic differential equation $$y'=y(b-ay) \, \textrm{with}\, a\neq 0, b\neq 0$$ has the non-trivial solution

$$y(t) = \frac{\frac{b}{a}}{1+c\cdot e^{-bt}}\tag{1}$$

$$\quad\quad = \frac{b}{(a+a\cdot c\cdot e^{-bt})}\tag{2}$$ where $c$ is a constant.

Why should we assume that $c$ is a positive real number?

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I tinkered a bit with the formatting, and while I was at is, multiplied numerator and denominator of $(1)$ by $a$ to get $(2)$. Feel free to "roll back" to your original post, if my edits are problematic to you. –  amWhy Nov 12 '12 at 15:10
    
There is actually no reason to assume that $c>0$. –  Artem Nov 12 '12 at 17:32

1 Answer 1

up vote 1 down vote accepted

There isn't. The logistic equation is commonly written in the form $$ {dP\over dt}=rP\left(1-{P\over K}\right), \quad P(0)=P_0, $$ and in the context of logistic population models,

  • $P$ is population
  • $t$ is time
  • $r$ is the intrinsic growth rate
  • $K$ is the carry capacity of the environment
  • $P_0$ is the initial population

Because of their physical meaning, each is taken to be positive. The solution is $$ P(t)={KP_0\over P_0+(K-P_0)e^{-rt}}={K\over 1+\left({K\over P_0}-1\right)e^{-rt}}. $$ This latter formulation matches the first form of your solution, just with $b:=r$ and $a:=r/K$, and $c:={K\over P_0}-1$.

There is no mathematical nor physical reason why we must have $c>0$. A negative value for $c$ would just mean that the initial population happened to be greater than the carrying capacity.

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