Modelling with differential equations (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

Modelling turns a worded rate description into a differential equation. "Rate proportional to amount" gives $\dfrac{dy}{dt} = ky$ (exponential growth or decay). "Rate proportional to the difference from a fixed value" gives Newton's law of cooling $\dfrac{dT}{dt} = -k(T - T_s)$ . "Limited growth toward a capacity" gives a logistic equation. Solve by separation, apply conditions, then interpret the long-term behaviour.

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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
Interpreting the logistic curve
Setting up from scratch
Exam-style questions

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:

Reading the rate phrase

"rate of change proportional to the amount present" gives $\dfrac{dy}{dt} = ky$ .
"rate proportional to the difference between $T$ and the surroundings $T_s$ " gives $\dfrac{dT}{dt} = -k(T - T_s)$ .
"growth limited by a carrying capacity $M$ " gives a logistic form such as $\dfrac{dP}{dt} = kP(M - P)$ .
The proportionality constant $k$ is positive; signs are set by whether the quantity rises or falls.

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 $T(t)$

Step 1: Write the differential equation, separate variables and integrate

Newton's law of cooling says the rate of change of temperature is proportional to the gap between the object's temperature and the surrounding temperature. With room temperature $20^{\circ}$ C the model is $\dfrac{dT}{dt} = -k(T - 20)$ . Separating variables and integrating both sides gives:

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

Step 2: Exponentiate and write the general solution

Exponentiating both sides removes the logarithm. Writing $A = e^{C}$ absorbs the integration constant into a single parameter.

$T - 20 = A e^{-kt}$ , so $T = 20 + A e^{-kt}$ .

Step 3: Apply the initial condition $T = 80$ at $t = 0$ to find $A$

Substituting $t = 0$ into the general solution immediately gives $A$ , because $e^{0} = 1$ .

$80 = 20 + A$ , so $A = 60$ . Thus $T = 20 + 60 e^{-kt}$ .

Step 4: Apply the second condition $T = 60$ at $t = 5$ to find $k$

The second piece of data gives an equation in $k$ alone, which we solve by taking a logarithm.

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

Step 5: Write the final model and interpret the long-term behaviour

Substituting the numerical value of $k$ gives the complete model. Examining what happens as time grows large confirms the physical meaning of the solution.

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 $P = \tfrac{M}{2}$ .

Always interpret the long-term behaviour

After solving, examine $t \to \infty$ . Decay and cooling models approach a limiting value; exponential growth is unbounded; logistic growth levels off at the capacity. SACE questions almost always ask you to state and interpret this limit.

Interpreting the logistic curve

The logistic solution $P = \dfrac{M}{1 + A e^{-kt}}$ has features SACE frequently asks you to read off. The carrying capacity $M$ is the horizontal asymptote the curve approaches as $t \to \infty$ . The constant $A$ is fixed by the initial population: substituting $t = 0$ gives $P(0) = \dfrac{M}{1 + A}$ , so $A = \dfrac{M - P(0)}{P(0)}$ . The growth rate is slow at first when $P$ is small, fastest at the inflection point $P = \tfrac{M}{2}$ , then slows again as $P$ nears $M$ . Being able to identify the capacity, the initial value, and the point of fastest growth from the formula is exactly what the interpretation marks reward.

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.

Common errors

Wrong sign on $k$: Cooling and decay need a negative rate, so the term is $-k(T - T_s)$ or $-kN$ with $k > 0$ . A sign error makes the model grow instead of decay.
Using only one data point: Most models have two unknowns ( $A$ and $k$ ); you need two conditions to find both.
Forgetting the surroundings term: Newton's cooling solution is $T = T_s + A e^{-kt}$ , not $A e^{-kt}$ alone; the $T_s$ offset is essential and sets the limiting temperature.
Skipping the interpretation: Stating the limiting value and what it means (room temperature, carrying capacity, equilibrium) is required for full marks.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style4 marks

Calculator-assumed. A butterfly population

B

grows by

\dfrac{dB}{dt} = 0.1B\left(1 - \dfrac{B}{K}\right)

, where

K

is a positive constant. You may assume

\dfrac{K}{B(K-B)} = \dfrac{1}{B} + \dfrac{1}{K-B}

. Show that

B = \dfrac{K}{1 + A e^{-0.1t}}

for some constant

A

Show worked answer →

Rewrite as $\dfrac{dB}{dt} = \dfrac{0.1B(K-B)}{K}$ and separate: $\dfrac{K}{B(K-B)}\,dB = 0.1\,dt$ . [1 mark]

Use the given partial fractions: $\int \left(\dfrac{1}{B} + \dfrac{1}{K-B}\right)dB = \int 0.1\,dt$ , giving $\ln|B| - \ln|K-B| = 0.1t + c$ , i.e. $\ln\dfrac{B}{K-B} = 0.1t + c$ . [1 mark]

Exponentiate: $\dfrac{B}{K-B} = Ce^{0.1t}$ . Take reciprocals: $\dfrac{K-B}{B} = \dfrac{1}{C}e^{-0.1t}$ , so $\dfrac{K}{B} - 1 = A e^{-0.1t}$ with $A = \tfrac{1}{C}$ . [1 mark]

Then $\dfrac{K}{B} = 1 + A e^{-0.1t}$ , so $B = \dfrac{K}{1 + A e^{-0.1t}}$ , as required. [1 mark]

SACE 2022-style5 marks

Calculator-assumed. The area covered by bacteria satisfies

\dfrac{dA}{dt} = \tfrac{1}{2}A\dfrac{(50 - A)}{50}

with

A(0) = 1

. Solve to show that

A = \dfrac{50}{1 + 49 e^{-0.5t}}

Show worked answer →

Separate: $\dfrac{50}{A(50-A)}\,dA = \tfrac{1}{2}\,dt$ . Partial fractions: $\dfrac{50}{A(50-A)} = \dfrac{1}{A} + \dfrac{1}{50-A}$ . [1 mark]

Integrate: $\ln|A| - \ln|50-A| = \tfrac{1}{2}t + c$ , so $\ln\dfrac{A}{50-A} = 0.5t + c$ . [1 mark]

Exponentiate: $\dfrac{A}{50-A} = Ce^{0.5t}$ . Apply $A(0) = 1$ : $\dfrac{1}{49} = C$ . [1 mark]

So $\dfrac{A}{50-A} = \tfrac{1}{49}e^{0.5t}$ , giving $49A = (50-A)e^{0.5t}$ , hence $A = \dfrac{50e^{0.5t}}{49 + e^{0.5t}}$ . [1 mark]

Divide through by $e^{0.5t}$ : $A = \dfrac{50}{1 + 49 e^{-0.5t}}$ , as required. [1 mark]

SACE 2021-style1 mark

Calculator-assumed. Given

A = \dfrac{50}{1 + 49 e^{-0.5t}}

models the area covered by bacteria, state the maximum area available for growth.

Show worked answer →

As $t \to \infty$ , $e^{-0.5t} \to 0$ , so the denominator $1 + 49e^{-0.5t} \to 1$ .

Therefore $A \to 50$ . The maximum (limiting) area is $50$ , the carrying capacity of the logistic model. [1 mark]

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board

What this dot point is asking

Translating words into a differential equation

Exponential growth and decay

Newton's law of cooling

A drink at 80∘80^{\circ}80∘C cools in a 20∘20^{\circ}20∘C room. After 555 minutes it is 60∘60^{\circ}60∘C. Find T(t)T(t)T(t)

Step 1: Write the differential equation, separate variables and integrate

Step 2: Exponentiate and write the general solution

Step 3: Apply the initial condition T=80T = 80T=80 at t=0t = 0t=0 to find AAA

Step 4: Apply the second condition T=60T = 60T=60 at t=5t = 5t=5 to find kkk

Step 5: Write the final model and interpret the long-term behaviour

Limited (logistic) growth

Interpreting the logistic curve

Setting up from scratch

Exam-style questions

Practise this

Sources & how we know this

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

Step 3: Apply the initial condition $T = 80$ at $t = 0$ to find $A$

Step 4: Apply the second condition $T = 60$ at $t = 5$ to find $k$