Logistic Models with Differential Equations

#7.9 Logistic Models with Differential Equations

Besides exponential models, differential equations can also be used with logistic models. In this topic, we’ll cover what logistic models are and how differential equations are relevant to them.

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#📈 Logistic Models with Differential Equations

The logistic model or logistic growth model is a differential equation that describes how a population grows over time—it grows proportionally to its size but stops growing when it reaches a certain size. Specifically, the model states that the rate of change of a population is jointly proportional to the size of the population and the difference between the population and the carrying capacity. The carrying capacity is the maximum number of individuals that the environment can sustain. 🌵

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Image of two graphs, one displaying exponential growth and the other displaying logistic growth with a carrying capacity.

Image Courtesy of Biology LibreTexts

Mathematically, we have the following differential equation, where $y$ is the population, $k$ is a positive constant representing the growth rate, and $M$ is the carrying capacity.

$$\frac{dy}{dt} = ky(M - y)$$

An equivalent form is the following…

$$\frac{dy}{dt} = ky(1 - \frac{y}{M})$$

The logistic growth model is widely used in ecology, biology, and other fields to describe the growth of populations of animals, plants, and microorganisms. It can also be used to describe the growth of human populations and other quantities that have a carrying capacity.

For the AP exam, you should be able to determine the carrying capacity of a logistic differential equation and the value of the dependent variable in a logistic differential equation at the point when it is changing fastest.

#✏️ Example of a Logistic Model

Let’s walk through an example problem to see how it’s done!

The population $P(t)$ of germs in a petri dish satisfies the following logistic differential equation where $t$ is measured in hours and the initial population is $400$.

$$\frac{dP}{dt} = 2P(5-\frac{P}{2000})$$

1. What is the carrying capacity of the population?

2. What is the population’s size when it’s growing the fastest?

#1) Solving for Carrying Capacity

When $t$ goes to infinity, $P(t)$ approaches the carrying capacity—there is a horizontal asymptote at the carrying capacity. This means that no matter the initial condition, the population will stabilize at the carrying capacity.

Additionally, $\frac{dP}{dt}$ approaches 0. So to find the carrying capacity, we just need to find $P$ such that $\frac{dP}{dt}$ approaches 0. $$0=2P(5-\frac{P}{2000})$$

There are two cases where $2P(5-\frac{P}{2000})$ equals 0—one where $2P$ equals $0$ (which just means $P$ is $0$) and one where $5-\frac{P}{2000}$ equals $0$.

$$0=5-\frac{P}{2000}$$

$$\frac{P}{2000}=5$$

$$P=10000$$

Thus, the carrying capacity is $10000.$

An alternative way to solve for the carrying capacity is to rewrite the equation in the form $\frac{dy}{dt} = ky(M - y)$ where $M$ is the carrying capacity.

We can rewrite $\frac{dP}{dt} = 2P(5-\frac{P}{2000})$ in this form by dividing $3P$ by $2000$ and multiply $5-\frac{5}{2000}$ by $2000$.

$$\frac{dP}{dt} = \frac{P}{1000}(10000-P)$$

Thus, we can see here that the carrying capacity is $10000.$

#2) Population Growth

The population’s size grows the fastest at the vertex of $\frac{dP}{dt}$ which is a downward quadratic equation with zeros at $P=0$ and the carrying capacity. Since the vertex is halfway between the two zeros, we just need to divide the carrying capacity by $2$. Therefore, the point of fastest change occurs at…

$$y=\frac{M}{2}$$

We already figured out in the previous part that the carrying capacity is $10,000$.

$$y=\frac{10,000}{2}$$

Thus, the population’s size when it’s growing the fastest is $5000$.

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#📝 Logistic Model Practice Problem

Now it’s time for you to do a practice problem! 🌟

The population $P(t)$ of bacteria in a petri dish satisfies the logistic differential equation $\frac{dP}{dt} = 4P(10-\frac{P}{2000})$ where $t$ is measured in hours and the initial population is $300$.

1. What is the carrying capacity of the population?

2. What is the population’s size when it’s growing the fastest?

#✅ Answers and Solutions

Make sure you give the question a try before scrolling!

#1) Solving for Carrying Capacity

When $t$ goes to infinity, $P(t)$ approaches the carrying capacity and $\frac{dP}{dt}$ approaches 0. So to find the carrying capacity, we just need to find $P$ such that $\frac{dP}{dt}$ approaches 0. $$0=4P(10-\frac{P}{2000})$$

There are two cases where $4P(10-\frac{P}{2000})$ equals 0—one where $4P$ equals $0$ (which just means $P$ is $0$) and one where $10-\frac{P}{2000}$ equals $0$.

$$0=10-\frac{P}{2000}$$

$$\frac{P}{2000}=10$$

$$P=20000$$

Thus, the carrying capacity is $20000$.

#2) Population Growth

Remember from the question we did together that the population’s size grows the fastest at…

$$y=\frac{M}{2}$$

We already figured out in the previous part that the carrying capacity is $20,000$, so all we have to do is plug in.

$$y=\frac{20,000}{2}$$

Thus, the population’s size when it’s growing the fastest is $10000$.

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#✨ Closing

Great job! Remember that the logistic growth model is a type of differential equation used to describe a growth process that is self-limiting. Here is a quick summary of the points we covered in this guide:

1. The term $ky$ indicates that the growth rate is proportional to the current population size.
2. The term $(1 - \frac{y}{M})$ implies that the growth rate decreases as $y$ approaches $M$, the carrying capacity.
3. When the initial population size $(y_0)$ is less than the carrying capacity ($y_0<M)$, the population will grow exponentially, as the population size is far from the carrying capacity. But, as $y$ approaches $M$, the growth rate will decrease. Think of it as you’re approaching the highest limit.
4. At $y=M$, the growth stops ($\frac{dy}{dt}=0$).
5. As $t$ approaches infinity, $y$ approaches $M$.
6. The point at which the population is changing the fastest is where $y$ is half the carrying capacity, $y=\frac{M}{2}$.

And…guess what? You made it to the end of unit seven! Here’s the start of unit eight if you want to get a head start. 🏃