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.

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?

share|improve this question
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer


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.