SASpecialist MathematicsSyllabus dot point

How do we translate a real situation into a differential equation and interpret its solution?

Formulate and solve differential equations modelling real situations such as growth, cooling and limited growth, and interpret the solutions.

Setting up differential equations from worded situations including exponential growth and decay, Newton's law of cooling and logistic limited growth, solving them, and interpreting the long-term behaviour.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Translating words into a differential equation
Exponential growth and decay
Newton's law of cooling
Limited (logistic) growth
Setting up from scratch

What this dot point is asking

You need to formulate differential equations from real situations, solve them, and interpret the resulting solutions including their long-term behaviour.

Translating words into a differential equation

The phrase describing the rate dictates the equation:

Exponential growth and decay

The simplest model, $\dfrac{dy}{dt} = ky$ , solves to $y = A e^{kt}$ (see the separable-equations method). Use two pieces of data: one to find $A$ , another to find $k$ . Interpretation: as $t \to \infty$ the quantity grows without bound if $k > 0$ or decays to $0$ if $k < 0$ .

Newton's law of cooling

A body at temperature $T$ in surroundings at $T_s$ cools at a rate proportional to the temperature difference:

\frac{dT}{dt} = -k(T - T_s).

Newton's law of cooling

A drink at $80^{\circ}$ C cools in a $20^{\circ}$ C room. After $5$ minutes it is $60^{\circ}$ C. Find $T(t)$

The model is $\dfrac{dT}{dt} = -k(T - 20)$ . Separate and integrate:

\int \frac{1}{T - 20}\,dT = \int -k\,dt \implies \ln|T - 20| = -kt + C.

Exponentiate:

T - 20 = A e^{-kt}

, so

T = 20 + A e^{-kt}

Initial condition $T = 80$ at $t = 0$ : $80 = 20 + A$ , so $A = 60$ . Thus $T = 20 + 60 e^{-kt}$ .

Second condition $T = 60$ at $t = 5$ :

60 = 20 + 60 e^{-5k} \implies 40 = 60 e^{-5k} \implies e^{-5k} = \tfrac{2}{3}.

-5k = \ln\tfrac{2}{3}

, giving

k = -\tfrac{1}{5}\ln\tfrac{2}{3} = \tfrac{1}{5}\ln\tfrac{3}{2} \approx 0.0811

The model is $T = 20 + 60 e^{-0.0811 t}$ . As $t \to \infty$ , $e^{-0.0811t} \to 0$ , so $T \to 20^{\circ}$ C - the drink approaches room temperature, as expected.

Limited (logistic) growth

When growth slows as a population nears a carrying capacity $M$ , the logistic equation $\dfrac{dP}{dt} = kP(M - P)$ applies. Its solution is an S-shaped curve that rises from a small initial value and levels off at $M$ . The key interpretation: $\dfrac{dP}{dt} \to 0$ as $P \to M$ , so the population stabilises at the carrying capacity, and the steepest growth occurs at the halfway point $P = \tfrac{M}{2}$ .

Setting up from scratch

A worked setup: a tank gains salt at a constant rate and loses it proportionally to the amount present, giving $\dfrac{dS}{dt} = a - bS$ . This is separable and solves to an exponential approaching the equilibrium $S = \tfrac{a}{b}$ , where the rate of change is zero. Recognising the equilibrium (where $\dfrac{dS}{dt} = 0$ ) often answers the long-term question without fully solving.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 SACE Stage 24 marksA butterfly population B has growth modelled by dB/dt = 0.1 B(1 - B/K), where K is a positive constant. You may assume K/(B(K-B)) = 1/B + 1/(K-B). Using integration techniques, show that the population can be modelled by B = K/(1 + A e^(-0.1t)) for some constant A.

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Rewrite the equation: dB/dt = 0.1 B(K-B)/K. Separate variables: K/(B(K-B)) dB = 0.1 dt. [1 mark]

Use the given partial fractions on the left: integral of (1/B + 1/(K-B)) dB = integral of 0.1 dt.
This gives ln|B| - ln|K-B| = 0.1 t + c, i.e. ln(B/(K-B)) = 0.1 t + c. [1 mark]

Exponentiate: B/(K-B) = e^(0.1t + c) = C e^(0.1t) for constant C. Take reciprocals: (K-B)/B = (1/C) e^(-0.1t), so K/B - 1 = A e^(-0.1t) where A = 1/C. [1 mark]

Then K/B = 1 + A e^(-0.1t), and rearranging gives B = K/(1 + A e^(-0.1t)), as required. [1 mark]

2017 SACE Stage 25 marksThe area covered by alpha bacteria satisfies dA/dt = (1/2) A((50 - A)/50) with initial condition A(0) = 1. Use integration to solve the differential equation and show that A = 50/(1 + 49 e^(-0.5t)).

Show worked answer →

Separate variables: 50/(A(50-A)) dA = (1/2) dt. Decompose the left by partial fractions: 50/(A(50-A)) = 1/A + 1/(50-A). [1 mark]

Integrate: integral of (1/A + 1/(50-A)) dA = integral of (1/2) dt, giving ln|A| - ln|50-A| = (1/2) t + c, so ln(A/(50-A)) = 0.5 t + c. [1 mark]

Exponentiate: A/(50-A) = C e^(0.5t). Apply A(0) = 1: 1/(50-1) = C, so C = 1/49. Thus A/(50-A) = (1/49) e^(0.5t). [1 mark]

Solve for A: 49 A = (50 - A) e^(0.5t). Rearranging, A(49 + e^(0.5t)) = 50 e^(0.5t), so A = 50 e^(0.5t)/(49 + e^(0.5t)). [1 mark]

Divide numerator and denominator by e^(0.5t): A = 50/(49 e^(-0.5t) + 1) = 50/(1 + 49 e^(-0.5t)), as required. [1 mark]

2017 SACE Stage 21 marksGiven A = 50/(1 + 49 e^(-0.5t)) models the area covered by alpha bacteria, state the maximum area that is available for bacterial growth.

Show worked answer →

As t increases without bound, e^(-0.5t) -> 0, so the denominator 1 + 49 e^(-0.5t) -> 1. [implicit]

Therefore A -> 50/(1) = 50. The maximum (limiting) area available for bacterial growth is 50 cm^2, which is the carrying capacity of the logistic model. [1 mark]