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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 $N^{'}$ 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^{'} = k \cdot N \cdot (K - N)

The general solution is given by the formula:

N = \frac{K}{C e^{- K k t} + 1}

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

\begin{array}{l} \int \frac{1}{N \cdot (K - N)} d N \\ = & \frac{1}{K} \cdot \ln \frac{N}{K - N} + C, 0 < N (0) < K \end{array}

\begin{array}{l} \int \frac{1}{N \cdot (K - N)} d N = \frac{1}{K} \cdot \ln \frac{N}{K - N} + C, 0 < N (0) < K . \end{array}

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 $N (0) = 800$ , $N^{'} = 320$ , $K = 7500$ . You find $k$ by entering the values into the equation and solving for $k$ : $\begin{array}{l} 320 & = k \cdot 800 \cdot (7500 - 800) \\ k & = 0.000 06 \end{array}$

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

N^{'} = 0.000 06 \cdot N \cdot (7500 - N)

Solve the differential equation by entering into the formula:

\begin{array}{l} N & = \frac{7500}{C e^{- 7500 \cdot 0.000 06 t} + 1} \\ = \frac{7500}{C e^{- 0.4478 t} + 1} \end{array}

N = \frac{7500}{C e^{- 7500 \cdot 0.000 06 t} + 1} = \frac{7500}{C e^{- 0.4478 t} + 1}

To find the growth model in this case you need to find

C

. From the text you know the initial condition

N (0) = 800

. Enter that into the equation for

N

and you get

\begin{array}{l} 800 & = \frac{7500}{C e^{- 0.4478 \cdot 0} + 1} \\ = \frac{7500}{C + 1} | \cdot (C + 1) \\ 800 \cdot (C + 1) & = 7500 \\ 800 C + 800 & = 7500 \\ C = \frac{7500 - 800}{800} & = 8.375 \end{array}

\begin{array}{l} 800 = \frac{7500}{C e^{- 0.4478 \cdot 0} + 1} & = \frac{7500}{C + 1} | \cdot (C + 1) \\ 800 \cdot (C + 1) & = 7500 \\ 800 C + 800 & = 7500 \\ C = \frac{7500 - 800}{800} & = 8.375 \end{array}

This gives the growth model:

N (t) = \frac{7500}{8.375 e^{- 0.4478 t} + 1}

You can now find the number of bacteria for $t = 5$ and $t = 13$ : $\begin{array}{l} N (5) & = \frac{7500}{8.375 e^{- 0.4478 \cdot 5} + 1} = 3963 \\ N (13) & = \frac{7500}{8.375 e^{- 0.4478 \cdot 13} + 1} = 7318 \end{array}$

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

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