Logistic Growth

Unit 7 · Differential Equations

Logistic Growth (BC Only)

The logistic DE dP/dt = kP(1 − P/M) models growth that accelerates early and slows as P approaches the carrying capacity M. The growth rate is maximum at P = M/2 (the inflection point). Below M/2 the curve is concave up; above M/2 it is concave down.

Logistic Differential Equation

\frac{dP}{dt} = kP\!\left(1 - \frac{P}{M}\right)

General solution

P(t) = \frac{M}{1 + Ae^{-kMt}}, \quad A = \frac{M - P_0}{P_0}

Key Features

Carrying capacity (equilibrium)

P = M

Maximum growth rate at

P = \frac{M}{2} \;(\text{inflection point})

Long-run behavior

P \to M \text{ as } t \to \infty

Caution: Do not confuse P = M/2 (where growth rate is fastest, inflection point) with P = M (carrying capacity, where growth stops).

Type 1

Analyzing Logistic Model Features

From the logistic DE, identify the carrying capacity, find the inflection point, and describe long-run behavior.

Example 1

A population grows with M = 500 and k = 0.004. (a) At what P is growth fastest? (b) If P(0) = 100, is the graph concave up or concave down at t = 0?

Example 2

Identify the equilibrium solutions of dP/dt = 0.002P(1 − P/800) and describe behavior when P > 800.

Type 2

Concavity and Inflection in Logistic Curves

The second derivative of P(t) determines concavity. The sign of d²P/dt² depends on whether P is above or below M/2.

Example 3

For dP/dt = kP(1 − P/M), show that the inflection point occurs at P = M/2 by finding d²P/dt².

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