Learn More  Fiveable Community students are already meeting new friends, starting study groups, and sharing tons of opportunities for other high schoolers. Soon the Fiveable Community will be on a totally new platform where you can share, save, and organize your learning links and lead study groups among other students! 🎉  # 7.9 Logistic Models with Differential Equations

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#differentialequations

written by jacob jeffries Logistic models describe phenomena using a logistic differential equation: 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: 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: 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: 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: Working with the L/3 case: Working with the L/2 case: Substituting Eq. 54 back into Eq. 48 yields a much easier integration: 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: 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.

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