A equals C minus P naught over P naught, or substitute the initial value at t equals zero.

Year 12 Maths Extension 1 (2027) Further applications of calculus

Logistic equation dP/dt = kP(1 − P/C)

Q: What is the logistic equation?

dP by dt equals k P times one minus P over C, modelling growth that levels off at a carrying capacity C.

Q: What is the logistic solution?

P equals C over one plus A e to the minus k t, with P approaching C as t grows.

Q: Where is growth fastest?

At P equals half the carrying capacity, the inflection point of the S-curve.

Q: Why use partial fractions?

To integrate one over P times C minus P after separating variables.

20 practice questions 2 video lessons Theory + worked examples

Explore the logistic equation in NSW Year 12 Mathematics Extension 1. It refines exponential growth by adding a carrying capacity, so a population grows quickly at first and then levels off.

You will learn to solve the logistic differential equation using partial fractions, interpret the resulting S-shaped curve, and identify the carrying capacity — a realistic growth model applied in biology, chemistry and economics in Extension 1.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

Modelling Population Growth: The Logistic Equation - Differential Equations Watch
Logistic Differential Equation (general solution) Watch

Theory

The logistic equation $\frac{d P}{d t} = k P (1 - \frac{P}{C})$ models growth that levels off at a carrying capacity $C$ , with solution $P = \frac{C}{1 + A e^{- k t}}$ . This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-05.

The logistic equation models growth that levels off at a carrying capacity $C$ : $\frac{d P}{d t} = k P (1 - \frac{P}{C})$ . Growth is rapid when $P$ is small and slows as $P \to C$ .

Solving it (by partial fractions and separation) gives $P = \frac{C}{1 + A e^{- k t}}$ . As $t \to \infty$ , $P \to C$ ; the constant $A = \frac{C - P_{0}}{P_{0}}$ is fixed by the initial value.

Growth is fastest at $P = \frac{C}{2}$ , the point of inflection of the $S$ -curve.

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01. Students decompose $\frac{1}{P (C - P)}$ into partial fractions to obtain the logistic function.

The

S

-curve rises fastest at

P = \frac{C}{2}

and levels off at

C

The growth rate

k P (1 - \frac{P}{C})

is greatest at

P = \frac{C}{2}

\frac{d P}{d t} = k P (1 - \frac{P}{C}) ⟹ P = \frac{C}{1 + A e^{- k t}} .

dP/dt = kP(1 - P/C) has solution P = C / (1 + A e^(-kt))

Use partial fractions before integrating:

\frac{1}{P (C - P)} = \frac{1}{C} (\frac{1}{P} + \frac{1}{C - P}), A = \frac{C - P_{0}}{P_{0}} .

partial fractions 1/(P(C-P)) = (1/C)(1/P + 1/(C-P)); A = (C - P0)/P0

Limiting value. $P \to C$ as $t \to \infty$ , and the maximum growth rate occurs at $P = \frac{C}{2}$ , not at $P = 0$ .

How to work with the logistic model

Read $C$ and $A$ : the carrying capacity is $C$ ; $A = \frac{C - P_{0}}{P_{0}}$ from the initial value.
To derive the solution, separate variables and use partial fractions on $\frac{1}{P (C - P)}$ .
Evaluate $P (t)$ by substituting $t$ into $P = \frac{C}{1 + A e^{- k t}}$ .
Interpret the limit $C$ and the fastest-growth point $\frac{C}{2}$ .

Example 1 — Read the model

For

P = \frac{600}{1 + 5 e^{- 0.4 t}}

, state the initial population and the carrying capacity.

Solution

$P (0)$	$=$	$\frac{600}{1 + 5} = 100$
$P (\infty)$	$=$	$600$

initial 100, carrying capacity 600

Initial $100$ , carrying capacity $600$ .

Example 2 — Find A

The solution is

P = \frac{C}{1 + A e^{- k t}}

with

C = 800

and

P (0) = 200

. Find

A

Solution

$200$	$=$	$\frac{800}{1 + A}$
$1 + A$	$=$	$4 \Rightarrow A = 3$

A = 3

$A = 3$ .

Example 3 — Derive the solution

Solve

\frac{d P}{d t} = k P (1 - \frac{P}{C})

Solution

Separate, then use partial fractions.

$\int (\frac{1}{P} + \frac{1}{C - P}) d P$	$=$	$\int k d t$
$\ln \| \frac{P}{C - P} \|$	$=$	$k t + c$

rearranges to P = C/(1 + A e^(-kt))

Rearranging: $P = \frac{C}{1 + A e^{- k t}}$ .

Example 4 — Evaluate

For

P = \frac{1000}{1 + 4 e^{- 0.2 t}}

, find

P

when

t = 10

Solution

$P (10)$	$=$	$\frac{1000}{1 + 4 e^{- 2}}$
	$\approx$	$\frac{1000}{1.541} \approx 649$

P(10) approx 649

$P (10) \approx 649$ .

Common pitfalls

Fastest growth point. Growth is fastest at

P = \frac{C}{2}

, not at

P = 0

Skipping partial fractions. Decompose

\frac{1}{P (C - P)}

before integrating.

Finding $A$ . Use

A = \frac{C - P_{0}}{P_{0}}

from the initial value, or substitute

t = 0

The limit. The carrying capacity

C

is the horizontal asymptote as

t \to \infty

Frequently asked questions

What is the logistic equation?

$\frac{d P}{d t} = k P (1 - \frac{P}{C})$ , modelling growth that levels off at a carrying capacity $C$ .

What is the logistic solution?

$P = \frac{C}{1 + A e^{- k t}}$ , with $P \to C$ as $t \to \infty$ .

Where is growth fastest?

At $P = \frac{C}{2}$ , the inflection point of the $S$ -curve.

How do you find A?

$A = \frac{C - P_{0}}{P_{0}}$ , or substitute the initial value at $t = 0$ .

Why use partial fractions?

To integrate $\frac{1}{P (C - P)}$ after separating variables.

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Logistic equation dP/dt = kP(1 − P/C)

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How to work with the logistic model

Common pitfalls

Frequently asked questions

Theory