7.9 Logistic Models with Differential Equations

BC only. The logistic model, dP/dt = kP(1 − P/L), describes growth that is nearly exponential at first and then levels off at a carrying capacity L. You should be able to read L, the long-run behavior and the point of fastest growth straight from the equation, without solving it.

Key terms

logistic growth
carrying capacity
dP/dt = kP(1 − P/L)
limiting value
point of fastest growth

The equation and its two forms

This topic is tested only on the BC exam.

Exponential growth can't go on forever: food, space or other limits slow it down. The logistic model builds that in. Its growth rate depends on two things multiplied together: how big P is, and how much room is left before the limit L. You'll see it written two ways:

dP/dt = kP(1 − P/L), or dP/dt = cP(L − P).

They're the same model: kP(1 − P/L) = (k/L)P(L − P), so c = k/L. In either form, L is the carrying capacity, the value P approaches in the long run. Here k > 0.

To find L from either form, ask which positive value of P makes dP/dt = 0. In the first form, that's the P that makes 1 − P/L zero. In the second, it's the P that makes L − P zero.

Reading the behavior

dP/dt = 0 when P = 0 or P = L. These are equilibrium solutions.
If 0 < P < L, then dP/dt > 0, so P increases toward L.
If P > L, then dP/dt < 0, so P decreases toward L.
So for any starting value P(0) > 0, lim (t → ∞) P(t) = L.
When P is small, P/L is close to 0, so dP/dt ≈ kP, which is almost exponential growth.

Where growth is fastest

dP/dt = kP − (k/L)P² is a downward-opening parabola in P, with zeros at P = 0 and P = L. Its maximum is halfway between, at P = L/2. So the population grows fastest when it's at half the carrying capacity.

You can also see this with concavity. Differentiate with respect to t: d²P/dt² = k(1 − 2P/L)·dP/dt. When 0 < P < L/2, both factors are positive, so the graph of P is concave up. When L/2 < P < L, it's concave down. The graph of P against t is an S-shaped curve with its inflection point at P = L/2.

The solution formula

Solving the equation with separation of variables and partial fractions (6.12) gives P(t) = L/(1 + Ae^(−kt)), where A = (L − P₀)/P₀. You mostly need to interpret the model rather than solve it, but recognizing this form helps on multiple-choice questions: L/(1 + Ae^(−kt)) → L as t → ∞.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Reading a logistic model
A population of fish satisfies dP/dt = 0.4P(1 − P/500), with P(0) = 50. (a) Find lim (t → ∞) P(t). (b) What is P when the population is growing fastest, and how fast is it growing then? (c) Is the graph of P concave up or down when P = 100?
Show the solutionHide the solution
1. Step 1: (a) The carrying capacity is L = 500. Since 0 < P(0) < 500, P increases toward 500. The limit is 500.
2. Step 2: (b) Fastest growth at P = L/2 = 250. Then dP/dt = 0.4(250)(1 − 250/500) = 0.4(250)(0.5) = 50 fish per unit of time.
3. Step 3: (c) P = 100 is below L/2 = 250, so the graph of P is concave up there (growth is still speeding up).
Answer: (a) 500 (b) P = 250, growing at 50 fish per unit time (c) Concave up
Example 2
Trap: the other form, and starting above L
A quantity y satisfies dy/dt = 0.002y(800 − y). (a) What is the carrying capacity? (b) If y(0) = 1000, describe y(t) for t ≥ 0. (c) If instead y(0) = 200, at what value of y is the quantity increasing fastest?
Show the solutionHide the solution
1. Step 1: (a) Rewrite: 0.002y(800 − y) = 1.6y(1 − y/800). So L = 800. (The tempting wrong answer is 0.002 or 1/0.002 = 500.)
2. Step 2: (b) y(0) = 1000 > 800, so dy/dt < 0. y decreases toward 800 and approaches it as t → ∞, never dropping below it.
3. Step 3: (c) Starting at 200 (between 0 and 800), y increases fastest at y = 800/2 = 400.
Answer: (a) 800 (b) y decreases and approaches 800 (c) y = 400

Common mistakes

Reading the carrying capacity from the wrong number in the cP(L − P) form. L is the value that makes the second factor zero.
Saying the population grows fastest at t = L/2. It grows fastest when P = L/2; the time when that happens is different.
Assuming P always increases. If P starts above L, it decreases toward L.
Forgetting the equilibrium P = 0: if P(0) = 0, P stays 0.

On the exam

Expect multiple-choice questions asking for the limit of P, the value of P where growth is fastest, or which differential equation describes a logistic situation.
If a free-response question gives a logistic equation, use the factored form to justify: “dP/dt > 0 for 0 < P < L, so P increases toward L.”

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