Logistic Differential Equations: Examples

A logistic differential equation, dP/dt = kP(1 − P/L), models growth that slows down as it nears a carrying capacity L. Its solution is P = L/(1 + Ae^(−kt)), an S-shaped curve. The population grows fastest when it reaches half the carrying capacity, P = L/2, and it approaches L over time.

Before this lesson: Exponential Models with Differential Equations, Integration by Partial Fractions and Logistic Functions

The idea: growth with a brake

Exponential growth, dP/dt = kP, never slows down. Real populations run out of food or space. The logistic equation multiplies kP by a brake factor, (1 − P/L).

When P is small, the brake factor is close to 1, so growth looks exponential. As P nears L, the factor shrinks toward 0, so growth stops. If P starts above L, the factor is negative, so the population falls back toward L.

L is the carrying capacity, the largest population the environment can support over time. k is the growth rate when the population is small.

Logistic differential equation formula

\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{L}\right) \quad\Longrightarrow\quad P(t) = \dfrac{L}{1 + A\,e^{-kt}}, \quad A = \dfrac{L - P_0}{P_0}

If P₀ is positive, P approaches L as t grows: the limit of P(t) is L.
The growth rate dP/dt is largest when P = L/2. That is the inflection point of the S-curve.
P = 0 and P = L are equilibrium (constant) solutions.
If P₀ is greater than L, A is negative and P decreases toward L.

Where the solution comes from

The equation is separable. The left side needs partial fractions.

\dfrac{L}{P(L - P)}\,dP = k\,dt

Write 1 − P/L as (L − P)/L, then separate the variables.

\left(\dfrac{1}{P} + \dfrac{1}{L - P}\right)dP = k\,dt

Split into partial fractions.

\ln|P| - \ln|L - P| = kt + C

Integrate both sides (see integrals that give natural logarithms).

\ln\left|\dfrac{P}{L - P}\right| = kt + C

Combine the logs (see properties of logarithms).

\dfrac{P}{L - P} = B\,e^{kt}

Exponentiate and let B = ±e^C.

\dfrac{L - P}{P} = A\,e^{-kt}

Take reciprocals of both sides, with A = 1/B.

\dfrac{L}{P} - 1 = A\,e^{-kt}

Split the fraction.

P = \dfrac{L}{1 + A\,e^{-kt}}

Add 1 to both sides, then take reciprocals and multiply by L.

At t = 0, the last line gives P₀ = L/(1 + A), so A = (L − P₀)/P₀.

Logistic differential equation examples

Example 1: Read the equation

For dP/dt = 0.4P(1 − P/500), find the growth constant, the carrying capacity, the population where growth is fastest, and the fastest growth rate.

k = 0.4, \quad L = 500

Match the equation to dP/dt = kP(1 − P/L).

P = \dfrac{L}{2} = 250

Growth is fastest at half the carrying capacity.

\dfrac{dP}{dt} = 0.4(250)\left(1 - \dfrac{250}{500}\right)

Substitute P = 250 into the equation.

\dfrac{dP}{dt} = 100(0.5) = 50

Multiply.

Answer: k = 0.4, L = 500, and growth is fastest at P = 250, where it is 50 per unit of time.

Example 2: Rewrite the equation first

Find k and L for dP/dt = 0.002P(300 − P).

300 - P = 300\left(1 - \dfrac{P}{300}\right)

Factor 300 out of 300 − P.

\dfrac{dP}{dt} = 0.002(300)P\left(1 - \dfrac{P}{300}\right)

Substitute into the equation.

\dfrac{dP}{dt} = 0.6P\left(1 - \dfrac{P}{300}\right)

Multiply 0.002 by 300.

Answer: k = 0.6 and L = 300. The number 0.002 is not k; it is k/L.

Example 3: Solve with an initial value

Solve dP/dt = 0.4P(1 − P/500) with P(0) = 50. Find P(5) and the time when growth is fastest.

A = \dfrac{500 - 50}{50} = 9

Use A = (L − P₀)/P₀.

P = \dfrac{500}{1 + 9e^{-0.4t}}

Substitute L = 500, A = 9 and k = 0.4.

P(5) = \dfrac{500}{1 + 9e^{-2}}

Substitute t = 5.

P(5) \approx \dfrac{500}{1 + 1.21802} \approx 225.43

Evaluate 9e^(−2) and divide.

250 = \dfrac{500}{1 + 9e^{-0.4t}}

Fastest growth is at P = 250.

1 + 9e^{-0.4t} = 2

Multiply both sides by the denominator and divide by 250.

e^{-0.4t} = \dfrac{1}{9}

Subtract 1 and divide by 9.

t = \dfrac{\ln 9}{0.4} \approx 5.49

Take the natural log and divide by −0.4.

Answer: P = 500/(1 + 9e^(−0.4t)), P(5) ≈ 225, and growth is fastest at t ≈ 5.49.

Graph of two solutions of the logistic equation dP/dt = 0.4P(1 − P/500) for t from 0 to 16. The blue S-shaped curve starts at (0, 50), rises slowly, then steeply, passes its inflection point at about (5.49, 250), and levels off just below the dashed carrying-capacity line P = 500. A second, green dashed curve starts at (0, 700) and falls toward 500. A dotted gray line marks P = 250, half the carrying capacity. — Below L = 500, the population rises in an S-shape and grows fastest at P = 250. Above L, it falls back toward 500.

Example 4: Fish in a lake (word problem)

A lake can support 2,000 fish. It is stocked with 400 fish, and after 2 years there are 800. Assuming logistic growth, find k and when the lake will hold 1,600 fish.

A = \dfrac{2000 - 400}{400} = 4

Use A = (L − P₀)/P₀.

800 = \dfrac{2000}{1 + 4e^{-2k}}

Substitute t = 2 and P = 800.

1 + 4e^{-2k} = 2.5

Multiply by the denominator and divide by 800.

e^{-2k} = 0.375

Subtract 1 and divide by 4.

k = \dfrac{\ln 0.375}{-2} \approx 0.4904

Take the natural log and divide by −2.

1600 = \dfrac{2000}{1 + 4e^{-kt}}

Set P = 1600.

1 + 4e^{-kt} = 1.25

Multiply by the denominator and divide by 1600.

e^{-kt} = 0.0625

Subtract 1 and divide by 4.

t = \dfrac{\ln 16}{k} \approx \dfrac{2.7726}{0.4904} \approx 5.65

Take the natural log, use −ln 0.0625 = ln 16, and divide by k.

Answer: k ≈ 0.49 per year, and the lake reaches 1,600 fish after about 5.7 years.

Example 5: Starting above the carrying capacity

Solve dP/dt = 0.4P(1 − P/500) with P(0) = 700. Is the population rising or falling, and what is its limit?

\dfrac{dP}{dt} = 0.4(700)\left(1 - \dfrac{700}{500}\right) = 280(-0.4) = -112

Substitute P = 700. The rate is negative.

A = \dfrac{500 - 700}{700} = -\dfrac{2}{7}

Use A = (L − P₀)/P₀.

P = \dfrac{500}{1 - \tfrac{2}{7}e^{-0.4t}}

Substitute L, A and k.

P(2) = \dfrac{500}{1 - \tfrac{2}{7}e^{-0.8}} \approx \dfrac{500}{0.87162} \approx 573.65

Substitute t = 2 and evaluate.

\lim_{t \to \infty} P(t) = \dfrac{500}{1 - 0} = 500

As t grows, e^(−0.4t) approaches 0.

Answer: P = 500/(1 − (2/7)e^(−0.4t)). The population falls (by 112 per unit of time at first), is about 574 at t = 2, and approaches 500.

Common mistakes

Mistake	Example of the error	Fix
Reading k from the wrong form	For dP/dt = 0.002P(300 − P), saying k = 0.002	Factor out L first: k = 0.002 · 300 = 0.6.
Flipping A	$A = \dfrac{P_0}{L - P_0}$	$A = \dfrac{L - P_0}{P_0}$ .
Fastest growth at the carrying capacity	“Growth is fastest at P = 500”	At P = L, growth is zero. It is fastest at P = L/2.
Mixing up P and t	“Growth is fastest at t = 250”	L/2 is a population. Solve P(t) = L/2 to find the time.
Assuming every population grows	Saying P rises when P₀ = 700 and L = 500	Above L, the brake factor is negative, so P falls.
Using e^(kt) in the denominator	$P = \dfrac{L}{1 + Ae^{kt}}$	The exponent is −kt, so the denominator shrinks to 1.

Related methods

Exponential models

For small P, the logistic equation acts like dP/dt = kP from exponential models with differential equations. The brake factor is what bends the curve into an S.

Logistic functions without calculus

Logistic functions in Precalculus use the same S-shaped formula. This lesson shows where the formula comes from.

Slope fields and Euler’s method

A slope field for a logistic equation shows flat slopes along P = 0 and P = L. Euler’s method can estimate logistic populations without solving the equation.

Other limited-growth models

Newton’s law of cooling also approaches a limiting value, but it never has an S-shape: it changes fastest at the start.

Practice problems

For dP/dt = 0.3P(1 − P/800), find k, L and the population where growth is fastest.
Find k and L for dP/dt = 0.01P(50 − P).
For dP/dt = 2P(1 − P/90) with P(0) = 10, find the limit of P(t) as t grows.
Solve dP/dt = 0.5P(1 − P/100) with P(0) = 20.
For problem 4, find P(4).
For problem 4, when does P reach 50?
Find the fastest growth rate for dP/dt = 0.2P(1 − P/1000).
A rumor spreads through a school of 1,200 students by dN/dt = 0.8N(1 − N/1200), with t in days. Six students start it. When have half the students heard it?
For dP/dt = 0.5P(1 − P/100) with P(0) = 150, is P increasing or decreasing at the start? What is its limit?
Show that for dP/dt = kP(1 − P/L), the second derivative is k²P(1 − P/L)(1 − 2P/L), and find where it is zero.

Show answers

k = 0.3, \quad L = 800

Match to dP/dt = kP(1 − P/L).

P = \dfrac{800}{2} = 400

Fastest growth at L/2.

Answer: k = 0.3, L = 800, fastest at P = 400

50 - P = 50\left(1 - \dfrac{P}{50}\right)

Factor out 50.

\dfrac{dP}{dt} = 0.5P\left(1 - \dfrac{P}{50}\right)

Multiply 0.01 by 50.

Answer: k = 0.5, L = 50

L = 90

Match the equation; P₀ = 10 is positive.

\lim_{t \to \infty} P(t) = L = 90

Logistic solutions approach L.

Answer: 90

A = \dfrac{100 - 20}{20} = 4

Use A = (L − P₀)/P₀.

P = \dfrac{100}{1 + 4e^{-0.5t}}

Substitute L, A and k.

Answer: P = 100/(1 + 4e^(−0.5t))

P(4) = \dfrac{100}{1 + 4e^{-2}}

Substitute t = 4.

P(4) \approx \dfrac{100}{1.54134} \approx 64.88

Evaluate and divide.

Answer: P(4) ≈ 64.9

1 + 4e^{-0.5t} = 2

Set P = 50, multiply by the denominator and divide by 50.

e^{-0.5t} = \dfrac{1}{4}

Subtract 1 and divide by 4.

t = \dfrac{\ln 4}{0.5} \approx 2.77

Take the natural log and divide by −0.5.

Answer: t ≈ 2.77

P = \dfrac{1000}{2} = 500

Fastest at L/2.

\dfrac{dP}{dt} = 0.2(500)(1 - 0.5) = 50

Substitute P = 500.

Answer: 50 per unit of time

A = \dfrac{1200 - 6}{6} = 199

Use A = (L − N₀)/N₀.

1 + 199e^{-0.8t} = 2

Set N = 600, multiply by the denominator and divide by 600.

e^{-0.8t} = \dfrac{1}{199}

Subtract 1 and divide by 199.

t = \dfrac{\ln 199}{0.8} \approx 6.62

Take the natural log and divide by −0.8.

Answer: after about 6.6 days

\dfrac{dP}{dt} = 0.5(150)\left(1 - \dfrac{150}{100}\right)

Substitute P = 150.

\dfrac{dP}{dt} = 75(-0.5) = -37.5

Multiply.

Answer: decreasing at first (rate −37.5), with limit 100

10.

\dfrac{dP}{dt} = kP - \dfrac{k}{L}P^2

Distribute kP.

\dfrac{d^2P}{dt^2} = k\dfrac{dP}{dt} - \dfrac{2k}{L}P\dfrac{dP}{dt}

Differentiate with the chain rule.

\dfrac{d^2P}{dt^2} = k\dfrac{dP}{dt}\left(1 - \dfrac{2P}{L}\right)

Factor out k dP/dt.

\dfrac{d^2P}{dt^2} = k^2P\left(1 - \dfrac{P}{L}\right)\left(1 - \dfrac{2P}{L}\right)

Substitute dP/dt = kP(1 − P/L).

Answer: it is zero at P = 0, P = L and P = L/2. The S-curve’s inflection point is at P = L/2

Logistic differential equation FAQ

What is the logistic differential equation?

It is dP/dt = kP(1 − P/L), where k is the growth constant and L is the carrying capacity. It models growth that is nearly exponential at first and then levels off at L.

What is the carrying capacity?

The carrying capacity L is the population the environment can support in the long run. Logistic solutions with a positive starting value approach L as time goes on.

When is a logistic population growing fastest?

At half the carrying capacity, P = L/2. There the second derivative is zero, so the S-curve has an inflection point.

What is the solution to the logistic differential equation?

P(t) = L/(1 + Ae^(−kt)), where A = (L − P₀)/P₀ and P₀ is the starting population.

Summary

Logistic equation: dP/dt = kP(1 − P/L), with carrying capacity L.
Solution: P = L/(1 + Ae^(−kt)) with A = (L − P₀)/P₀.
P approaches L; P = 0 and P = L are equilibrium solutions.
Growth is fastest at P = L/2, the inflection point.
Factor the equation into the form kP(1 − P/L) before reading k and L.
A population above L decreases toward L.

Previous lesson: Newton’s Law of Cooling

Next lesson: Derivatives of Parametric Equations

Back to the Calculus 2 Index