# How to Find the Logistic Growth Using the Carrying Capacity of a Population

The carrying capacity $K$ is the maximum population of a particular group that may be sustained in a region. Logistic growth is when the growth factor ${N}^{\prime }$ is proportional to the population itself $N$ and proportional to the difference between the population $N$ and the carrying capacity $K$. You can therefore write logistic growth as a separable differential equation. The differential equation is:

 ${N}^{\prime }=k\cdot N\cdot \left(K-N\right)$

The general solution is given by the formula:

 $N=\frac{K}{C{e}^{-Kkt}+1}$

If you are asked to solve the integral, this equation is useful:

$\begin{array}{llll}\hfill & \int \frac{1}{N\cdot \left(K-N\right)}\phantom{\rule{0.17em}{0ex}}dN\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill =& \frac{1}{K}\cdot \mathrm{ln}\frac{N}{K-N}+C,\phantom{\rule{1em}{0ex}}0

$\begin{array}{lll}\hfill \int \frac{1}{N\cdot \left(K-N\right)}\phantom{\rule{0.17em}{0ex}}dN=\frac{1}{K}\cdot \mathrm{ln}\frac{N}{K-N}+C,\phantom{\rule{1em}{0ex}}0

Example 1

The number of bacteria in a polluted drinking water reservoir was originally 800. In the first hour afterwards, the number of bacteria increased by 320. Assume that the number of bacteria follows a model of logistic growth and that the carrying capacity is 7500.

Let $N$ be the number of bacteria after $t$ hours. Find $k$, determine the growth model and find how many bacteria there were after $5$ and $13$ hours.

From the text you see that $N\left(0\right)=800$, ${N}^{\prime }=320$, $K=7500$. You find $k$ by entering the values into the equation and solving for $k$: $\begin{array}{llll}\hfill 320& =k\cdot 800\cdot \left(7500-800\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =0.000\phantom{\rule{0.17em}{0ex}}06\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can now enter $k$ straight into the differential equation and solve this to find the growth model:

 ${N}^{\prime }=0.000\phantom{\rule{0.17em}{0ex}}06\cdot N\cdot \left(7500-N\right)$

Solve the differential equation by entering into the formula:

$\begin{array}{llll}\hfill N& =\frac{7500}{C{e}^{-7500\cdot 0.000\phantom{\rule{0.17em}{0ex}}06t}+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7500}{C{e}^{-0.4478t}+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $N=\frac{7500}{C{e}^{-7500\cdot 0.000\phantom{\rule{0.17em}{0ex}}06t}+1}=\frac{7500}{C{e}^{-0.4478t}+1}$

To find the growth model in this case you need to find $C$. From the text you know the initial condition $N\left(0\right)=800$. Enter that into the equation for $N$ and you get
$\begin{array}{llll}\hfill 800& =\frac{7500}{C{e}^{-0.4478\cdot 0}+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7500}{C+1}\phantom{\rule{1em}{0ex}}|\cdot \left(C+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 800\cdot \left(C+1\right)& =7500\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 800C+800& =7500\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill C=\frac{7500-800}{800}& =8.375\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill 800=\frac{7500}{C{e}^{-0.4478\cdot 0}+1}& =\frac{7500}{C+1}\phantom{\rule{2em}{0ex}}|\cdot \left(C+1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 800\cdot \left(C+1\right)& =7500\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 800C+800& =7500\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill C=\frac{7500-800}{800}& =8.375\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This gives the growth model:
 $N\left(t\right)=\frac{7500}{8.375{e}^{-0.4478t}+1}$

You can now find the number of bacteria for $t=5$ and $t=13$: $\begin{array}{llll}\hfill N\left(5\right)& =\frac{7500}{8.375{e}^{-0.4478\cdot 5}+1}=3963\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill N\left(13\right)& =\frac{7500}{8.375{e}^{-0.4478\cdot 13}+1}=7318\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The number of bacteria after 5 and 13 hours is $3963$ and $7318$ respectively.