Mathematics 9709 · syllabus 3.8
Cambridge A Level Mathematics 9709 — Differential equations topical past papers
Forming first-order differential equations from rates of change, solving them by separating the variables, and interpreting the solution in context.
- Questions
- 107
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 31
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What Differential equations covers
A differential equation links a quantity to its rate of change. In Paper 3 (section 3.8) you meet first-order equations whose variables can be separated, so that every y term sits with dy and every x term with dx, as in dy/dx = xy² or dN/dt = −kN. Integrating both sides gives a general solution with one arbitrary constant, and an initial condition fixes that constant to give a particular solution.
Questions are often set in context: a population, the temperature of a cooling liquid, the depth of water in a tank. You may have to form the equation yourself from a sentence such as 'the rate of decrease of x is proportional to the square root of x', introducing a constant of proportionality and finding its value from given data. A final part usually asks what the solution says, such as what happens in the long term or when the tank is empty.
Differential equations appear in Paper 3 only, and this is where its integration techniques are used in earnest: partial fractions, integration by parts, ∫f′(x)/f(x) dx, trigonometric identities and a given substitution can all be needed for one side. The laws of logarithms are then used to make y the subject. The exponential models that come out link back to section 3.2, and the idea of a rate of change goes back to connected rates of change in Paper 1.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 3.8 Differential equations
What you should be able to do
- Form a differential equation from a statement about a rate of change, with a constant of proportionality
- Separate the variables and integrate both sides
- Use an initial condition to find the constant of integration
- Rearrange the solution to give y explicitly in terms of x when asked
- Interpret the solution in context, including its long-term behaviour
Key ideas
- If dy/dx = f(x)g(y), then ∫ 1/g(y) dy = ∫ f(x) dx
- 'The rate of change of x' means dx/dt, and 'is proportional to' introduces a constant k
- ∫ 1/y dy = ln|y| + c, so dN/dt = kN has the general solution N = Ae^(kt)
- The general solution has one arbitrary constant, and one condition fixes it
How Differential equations has been examined
From 2022 to 2026, Differential equations was on Paper 3 · Pure Mathematics 3 in 11 of 11 exam series, worth about 11% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 3 · Pure Mathematics 3 | 31 | 11% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 8.
Filed in Momentum under: Differential Equation.
A worked example
The temperature θ °C of a liquid t minutes after it is set down satisfies dθ/dt = −k(θ − 20), where k is a positive constant. Initially θ = 80, and θ = 50 when t = 5. Find θ in terms of t and describe what happens to θ as t becomes large.
- Separate the variables: ∫ 1/(θ − 20) dθ = ∫ −k dt.
- Integrate: ln(θ − 20) = −kt + c, since θ − 20 > 0 throughout.
- At t = 0, θ = 80, so c = ln 60 and θ − 20 = 60e^(−kt).
- At t = 5, θ = 50: 30 = 60e^(−5k), so e^(−5k) = ½ and k = (ln 2)/5 = 0.139.
- So θ = 20 + 60e^(−0.139t); as t increases, e^(−0.139t) tends to 0.
Answer: θ = 20 + 60e^(−0.139t), where 0.139 = (ln 2)/5; θ approaches 20 °C
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Adding the constant of integration after rearranging rather than straight after integrating: ln y = x² + c gives y = Ae^(x²), not y = e^(x²) + c.
- Leaving dx or dy on the wrong side when separating, or integrating with respect to the wrong variable.
- Writing dx/dt = kx for a quantity that decreases: 'decreases at a rate proportional to x' needs dx/dt = −kx with k > 0.
- Turning ln y = ln x + c into y = x + c: the constant becomes a multiplier, so y = Ax.
Exam technique
- Put + c on one side as soon as you integrate, and substitute the initial condition before rearranging; it is usually easier.
- Show every step of the integration on each side, using partial fractions or integration by parts where one side needs them.
- Answer interpretation parts in a sentence about the context, such as 'the temperature approaches 20 °C'.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.