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7.9 Logistic Models with Differential Equations

4 min readjanuary 25, 2023

Jed Quiaoit

Jed Quiaoit

Jacob Jeffries

Jacob Jeffries

Jed Quiaoit

Jed Quiaoit

Jacob Jeffries

Jacob Jeffries


AP Calculus AB/BC ♾️

279 resources
See Units

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The logistic growth model is a mathematical model that describes how a population grows over time. It is based on the statement 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 can be sustained by the environment. 🍃
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.
The logistic growth model also has some important implications for sustainable development and resource management. It shows that populations cannot continue to grow indefinitely and that there is a limit to the resources that can sustain a population. Therefore, the model highlights the need for sustainable resource use and the importance of controlling population growth.

Logistic Differential Equations

Logistic models describe phenomena using a logistic differential equation:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1004).png?alt=media&token=a06cf19f-87d1-40ba-bf0f-76bed2f2c83f
The kL term is often grouped together into a single constant.
The most popular (no pun intended) model using a logistic differential equation is modeling population dynamics. This is because it accounts for exponential growth as well as a population capacity, as seen in Fig. 1.1.
There are some properties we are going to derive that are necessary for the AP exam, in particular, the multiple-choice section.
The first one is the solution to the differential equation. It is separable but in a strange way. We will use the form of the differential equation as presented in Eq. 46: 🦒
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1001).png?alt=media&token=ca7db6cb-9da7-4ad0-895d-0cc84fc45f74
This means we have to do a tough integral. As it is, we cannot integrate the left-hand side with normal techniques. We will assume that the function can be written in the following form:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1000).png?alt=media&token=a46d7cf1-c194-4c13-8548-e8537c59ea2c
This means the equation must be true for any value of f, given the value of f falls within the domain of the function.
The domain is of the function on the left-hand side in Eq. 47 is the following:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(999).png?alt=media&token=d382533f-6a42-4231-888c-b9a16902dd74
Which means we can pick any values of f except 0 and 1 to set up a system of equations to solve for A and B. Let’s choose L/3 and L/2:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(996).png?alt=media&token=71db7a92-87d6-425e-866a-7ff7ea8c6f1e
Working with the L/3 case:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(995).png?alt=media&token=f757c677-0f9d-46a6-8347-3ed8e37ede44
Working with the L/2 case:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(993).png?alt=media&token=5503186a-2aac-4ad8-923d-8a22b7844334
Substituting Eq. 54 back into Eq. 48 yields a much easier integration:
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The second one is the two relevant maxima: the maximum value of f and the maximum value of df/dx.
The maximum value of f (fmax) is a simple principle: it is simply L*. The best way to visualize this is by graphing the derived solution and seeing what happens as x gets infinitely large. In this case the graph approaches a “carrying capacity,'' which is expressed as L in Eq. 45.
The second value is a bit more difficult to solve. Let’s use implicit differentiation to find the second derivative of f using the form of the equation in Eq. 45: 📦
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(989).png?alt=media&token=ea8f8653-aad0-418c-a9ee-953ba74efa4e
One can solve for f to get fmax = L/2.
Of course, this isn’t the full story, as this just means L/2 is a critical point. You can verify this by finding the value of f’’’ to check that it is positive at L/2.

Footnotes

*This is not a maximum in the strict definition of the word; the function does not have a maximum but rather grows monotonically.

(1) Regarding Interpretations

The logistic differential equation and initial conditions can be interpreted without solving the differential equation, by analyzing the properties of the equation and the initial conditions. 😳
The logistic differential equation dy/dt = ky(a - y) describes how the population size changes over time. The term ky(a-y) represents the rate of change of the population, where k is the growth rate and a is the carrying capacity. The term ky represents the growth rate when the population is small, while the term -ky(y-a) represents the decline rate when the population is close to the carrying capacity.
The initial conditions, represented by y(0) = y0, specify the initial population size. The population size at time t = 0 is y0. By analyzing the initial conditions, we can understand the starting point of the population growth.
We can also understand the behavior of the population over time without solving the differential equation. For example, if the initial population size is y0 < a, then the population will grow exponentially, as the population size is far from the carrying capacity. On the other hand, if y0 > a, then the population will decline as the population size is close to the carrying capacity.
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