7.9 - Logistic Models with Differential Equations (BC ONLY)

Note: this is a BC Calculus topic. It’s not on the AB exam.

Introduction

Exponential growth increases forever with no ceiling. Logistic growth has a limiting value. The quantity grows like an exponential at first (faster and faster), then slows down as it approaches a maximum sustainable level.

7.9 - Logistic Models with Differential Equations (BC ONLY)

Key Features

Carrying capacity ( $L$ ): the maximum sustainable value. Also called the limiting value or ceiling.

Point of inflection: located at

y = \frac{L}{2}

, exactly halfway to the carrying capacity.

Before the inflection point: growth rate is increasing (concave up)

After the inflection point: growth rate is decreasing (concave down)

At the inflection point: growth rate is at its maximum

Important distinction: in the S-shaped curve above, the population keeps increasing the whole time. It’s the rate of increase that rises and then falls.

The Two Forms

A logistic differential equation can be written two equivalent ways.

Form 1:

\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right)

Form 2:

\frac{dy}{dt} = \frac{k}{L} \cdot y(L - y)

Converting between them: factor

\frac{1}{L}

out of Form 1.

ky\left(1 - \frac{y}{L}\right) = k \cdot \frac{1}{L} \cdot y \cdot (L - y) = \frac{k}{L} \cdot y(L-y)

Recognition Tip

The dependent variable appears twice. That’s the giveaway.

$\frac{dP}{dt} = 0.4P\left(1 - \frac{P}{500}\right)$ is logistic with $L = 500$

$\frac{dN}{dt} = 0.02N(80 - N)$ is logistic with $L = 80$

Finding Key Values

Example 1: Form with $(1 - y/L)$

Given

\frac{dy}{dt} = 0.5y\left(1 - \frac{y}{8}\right)

, find the carrying capacity and the inflection point.

This is Form 1, so

L

is the denominator under

y

L = 8

The inflection point is at

y = \frac{L}{2} = 4

Example 2: Form with $(L - y)$

Given

\frac{dP}{dt} = \frac{0.03}{200}P(200 - P)

, find the same things.

This is Form 2, so

L

is the number in the

(L - y)

factor:

L = 200

Maximum growth rate happens at

P = \frac{200}{2} = 100

Maximum Value vs. Maximum Rate

These get confused constantly.

Question	Answer	Why
Maximum value of the function	$y = L$	That’s the carrying capacity, the ceiling (when the solution starts below it)
Maximum rate of change	$y = \frac{L}{2}$	That’s the inflection point

Example 3

A population follows

\frac{dN}{dt} = 0.08N\left(1 - \frac{N}{1200}\right)

(a) Maximum population? That’s the carrying capacity:

L = 1200

(b) Population where growth rate is greatest? That’s the inflection point:

N = \frac{1200}{2} = 600

Showing It with the Second Derivative

Quick way to see it:

\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right)

is a downward-opening parabola in

y

, with zeros at

y = 0

and

y = L

. A parabola’s highest point is halfway between its zeros, at

y = \frac{L}{2}

On free response, show it with

\frac{d^2y}{dt^2}

. Expand, then differentiate with respect to

t

(remember

y

is a function of

t

, so each

y

term gets a

\frac{dy}{dt}

\frac{dy}{dt} = ky - \frac{k}{L}y^2 \quad \Rightarrow \quad \frac{d^2y}{dt^2} = k\frac{dy}{dt} - \frac{2k}{L}y\frac{dy}{dt} = k\left(1 - \frac{2y}{L}\right)\frac{dy}{dt}

For

0 < y < L

\frac{dy}{dt}

is positive, so the sign of

\frac{d^2y}{dt^2}

comes from

1 - \frac{2y}{L}

$y < \frac{L}{2}$ : $\frac{d^2y}{dt^2} > 0$ , so $\frac{dy}{dt}$ is increasing (concave up)

$y > \frac{L}{2}$ : $\frac{d^2y}{dt^2} < 0$ , so $\frac{dy}{dt}$ is decreasing (concave down)

\frac{dy}{dt}

is greatest at

y = \frac{L}{2}

. In Example 3,

\frac{d^2N}{dt^2} = 0.08\left(1 - \frac{N}{600}\right)\frac{dN}{dt}

changes sign at

N = 600

, and the fastest growth rate is

0.08(600)\left(1 - \frac{600}{1200}\right) = 24

Converting to Standard Form

Sometimes you have to do algebra to find

L

Example 4

Find the carrying capacity for

\frac{dy}{dt} = \frac{2}{3}y\left(5 - \frac{y}{4}\right)

Step 1: Expand.

\frac{dy}{dt} = \frac{10}{3}y - \frac{y^2}{6}

Step 2: Factor out the coefficient of $y$ to match Form 1.

= \frac{10}{3}y\left(1 - \frac{y}{20}\right)

(Check:

\frac{10}{3}y \cdot \frac{y}{20} = \frac{y^2}{6}

, which matches.)

Step 3: Identify $L$ .

L = 20

Example 5

Find the carrying capacity for

\frac{dy}{dt} = \frac{3}{4}y\left(7 - \frac{y}{20}\right)

Step 1: Expand.

\frac{dy}{dt} = \frac{21}{4}y - \frac{3y^2}{80}

Step 2: Factor out $\frac{21}{4}y$ .

= \frac{21}{4}y\left(1 - \frac{3y}{80} \cdot \frac{4}{21}\right) = \frac{21}{4}y\left(1 - \frac{y}{140}\right)

Step 3:

L = 140

Example 6: Word Problem (State Park Visitors)

Problem: The number of people

P

in a state park increases at a rate proportional to the product of the current number of people and the difference between the park’s capacity and the current number. Capacity is 3000. When there are 800 people, the rate of increase is 120 people per hour. Write a differential equation.

Step 1: Identify the model. “Rate proportional to the product of the current number and the difference from capacity” is a logistic model.

You’ll also see this worded as “jointly proportional to

P

and

3000 - P

.” Jointly proportional means proportional to the product, so it’s the same logistic model.

Step 2: Use Form 2.

\frac{dP}{dt} = \frac{k}{L} \cdot P(L - P)

Step 3: Plug in $L = 3000$ .

\frac{dP}{dt} = \frac{k}{3000} \cdot P(3000 - P)

Step 4: Find $k$ . When

P = 800

\frac{dP}{dt} = 120

120 = \frac{k}{3000} \cdot 800(2200) = \frac{1{,}760{,}000k}{3000}

k = \frac{120 \cdot 3000}{1{,}760{,}000} = \frac{360{,}000}{1{,}760{,}000} = \frac{9}{44}

Step 5: Write the equation.

\frac{dP}{dt} = \frac{9}{132{,}000} \cdot P(3000 - P)

Or in Form 1:

\frac{dP}{dt} = \frac{9}{44}P\left(1 - \frac{P}{3000}\right)

Using the Initial Condition

You can describe a logistic solution without solving the differential equation. Plug the starting value into the right side to get the sign of

\frac{dy}{dt}

, then compare the starting value to

L

and

\frac{L}{2}

Example 7: A lake is stocked with fish, and the number of fish

P

follows

\frac{dP}{dt} = 0.3P\left(1 - \frac{P}{800}\right)

where

t

is in years. The carrying capacity is

L = 800

, so the population grows fastest when

P = 400

If $P(0) = 100$ : $\frac{dP}{dt} = 0.3(100)\left(1 - \frac{100}{800}\right) = 26.25 > 0$ , so the population increases toward $800$ . It’s concave up until $P = 400$ , where it grows fastest, then concave down.

If $P(0) = 600$ : $\frac{dP}{dt} = 0.3(600)\left(1 - \frac{600}{800}\right) = 45 > 0$ , so it also increases toward $800$ . It starts above $400$ , so it’s already past its fastest growth: concave down the whole time, with no inflection point.

If $P(0) = 1000$ : $\frac{dP}{dt} = 0.3(1000)\left(1 - \frac{1000}{800}\right) = -75 < 0$ . The lake is overstocked, so the population decreases toward $800$ (concave up) and never gets near $400$ .

If $P(0) = 0$ or $P(0) = 800$ : $\frac{dP}{dt} = 0$ , so the population never changes. $P = 0$ and $P = 800$ are the equilibrium solutions.

In every case that starts above

0

, the limiting value is the carrying capacity:

\lim_{t \to \infty} P(t) = 800

So for “what happens in the long run?” questions, the answer is

L

no matter where a positive population starts. The initial condition only decides whether it rises or falls to get there, and whether it passes the fastest-growth point

\frac{L}{2}

on the way.

Solving the Logistic Equation (Optional)

The general solution to

\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right)

y = \frac{L}{1 + Ce^{-kt}}

(If the equation is written as

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

, with no

\frac{1}{L}

in front, the exponent becomes

-kLt

instead.)

where

L

is the carrying capacity,

k

is the growth constant,

C

comes from the initial condition, and

t

is time.

The derivation involves separating variables, doing partial fraction decomposition on

\frac{1}{y(1 - y/L)}

, integrating, and solving for

y

For most AP problems you don’t need this. Focus on pulling

L

and the inflection point straight out of the differential equation.

Strategy Summary

For multiple choice:

Spot the logistic form (dependent variable appears twice)

Find $L$ fast: in Form 1 it’s the denominator under $y$ ; in Form 2 it’s the number in $(L - y)$

Maximum value is $y = L$ (when $y$ starts below $L$ )

Maximum rate is at $y = \frac{L}{2}$

For free response:

Show work for the inflection point: set $\frac{d^2y}{dt^2} = 0$ and solve

Verify the concavity change by checking the sign of $\frac{d^2y}{dt^2}$ on both sides

State conclusions clearly

Common Question Types

Direct identification: “What is the carrying capacity?” Read $L$ off directly.

Inflection point: “At what population is growth fastest?” Answer $y = \frac{L}{2}$ .

Setting up: “Write a differential equation for…” Use context to spot logistic, then find $L$ and $k$ .

Manipulation required: The equation needs factoring first to get to standard form.

Key Takeaways

Logistic growth has a carrying capacity $L$

Two equivalent forms: $\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right)$ and $\frac{dy}{dt} = \frac{k}{L}y(L-y)$

Recognition: the dependent variable shows up twice

Maximum value is $L$ (when $y$ starts below $L$ )

Long run: $\lim_{t \to \infty} y = L$ for any starting value above $0$ . If it starts above $L$ , the solution decreases to $L$

Point of inflection is always at $y = \frac{L}{2}$

Maximum growth rate happens at the inflection point, NOT at the maximum value

On free response, show it with $\frac{d^2y}{dt^2} = k\left(1 - \frac{2y}{L}\right)\frac{dy}{dt}$ , which changes sign at $y = \frac{L}{2}$

Before the inflection point: concave up, growth rate rising

After the inflection point: concave down, growth rate falling

BC only, not on the AB exam

Maximum value ( $L$ ) is not the same as maximum rate ( $L/2$ )

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