Learn7357Paper 3 · Pure and StatisticsPure content (sections A–J)Modelling with differential equations

Pure content (sections A–J) · Subtopic 63 of 66

Modelling with Differential Equations

Populations, cooling drinks, draining tanks and spreading rumours. Solve the equation, use two pieces of data to fix the constants, and then ask the real questions: when, how much, and what happens in the long run?

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 7 · updated 2026-10-27

Section 1 of 14

Learning objectives

Solve growth, decay and cooling models and use two data points to find both constants.
Answer "when" questions by solving an exponential or logarithmic equation.
Solve logistic models of the form $\dfrac{dN}{dt}=kN(a-N)$ using partial fractions.
Describe long-term behaviour and comment on a model's limitations.

Pause and think

A population of rabbits grows according to $\dfrac{dN}{dt}=0.5N$ . What does the model predict after $50$ years? Is that believable?

Section 2 of 14

Core idea

Start hereWhen will the tea be drinkable?

Tea at $90$ °C in a $20$ °C room is $55$ °C after $10$ minutes. When will it reach $30$ °C? Newton's law of cooling, a differential equation, plus two data points answers it: about $28$ minutes.

The big idea

Modelling cycle: form the equation, solve it, use data to find constants, predict, then interpret and criticise (long-term behaviour, limitations).

1. Newton's law of cooling

Cooling from 90 °C in a 20 °C room

k = 0.04

A bigger

k

means faster cooling (a thin cup, a draught). In every case the drink approaches

20^\circ

C but never goes below it.

Worked exampleThe tea

Find when the tea reaches $30$ °C.

Goal: Time.
Strategy: Separate, use $t=0$ then the 10-min reading.

$\frac{d\theta}{dt}=-k(\theta-20)$
Model.
$\int\frac{d\theta}{\theta-20}=-\int k\,dt\Rightarrow\ln(\theta-20)=-kt+c$
Separate; integrate.
$t=0:\ c=\ln70\Rightarrow\theta=20+70e^{-kt}$
Initial condition.
$35=70e^{-10k}\Rightarrow k=\tfrac{\ln2}{10}=0.0693$
Data at $t=10$ .
$10=70e^{-kt}\Rightarrow t=\frac{\ln7}{k}=28.1$
Solve for $t$ .

AnswerAbout

28

minutes.

✓ Check

\theta\to20

t\to\infty

(room temperature) ✓.

2. Interpreting

Worked examplePopulation model

$\frac{dP}{dt}=0.02P$ , $P(0)=50$ (thousands). Find the doubling time and criticise the model.

Goal: Time and comment.
Strategy: Solve; set $P=100$ .

$P=50e^{0.02t}$
Solution.
$e^{0.02t}=2\Rightarrow t=\frac{\ln2}{0.02}=34.7$
Doubling.

Answer

34.7

years; but the model predicts unlimited growth, unrealistic (food, space).

✓ CheckLogistic models fix this ✓.

The modelling cycle

Drawing diagram…

If predictions disagree with data, revise the assumptions and go round again.

Pause and think

What does a larger $k$ mean physically for cooling?

Summary

Form → solve → constants → predict → criticise.

Model: A simplified mathematical description.

Newton’s law of cooling:

\theta = 20 + 70e^{-0.1t}

The solution of

\dfrac{d\theta}{dt}=-k(\theta-20)

approaches room temperature but never reaches it: the rate of cooling falls as the difference falls.

HistoryLimits to growth. In 1798 Thomas Malthus warned that populations grow exponentially while food supplies do not. In 1838 the Belgian mathematician Pierre-François Verhulst answered with the logistic equation, in which growth slows as a population approaches the most its environment can support. His model was ignored for nearly a century before being rediscovered.

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