What are differential equations and why are they important in Pure Mathematics 3 for Cambridge International A Levels?

Differential equations are mathematical equations that involve functions and their derivatives. They are crucial in Pure Mathematics 3 as they model real-world phenomena where rates of change are involved, such as in physics, engineering, and economics. Understanding differential equations helps students develop problem-solving skills and apply mathematical concepts to practical situations.

How do you solve a first-order differential equation in the Cambridge International A Levels curriculum?

To solve a first-order differential equation, you typically use methods such as separation of variables or integrating factors. In separation of variables, you rearrange the equation to isolate the variables on different sides and then integrate both sides. For integrating factors, you multiply the entire equation by a specific function to make it integrable. These methods are essential for solving problems in Pure Mathematics 3.

What is the difference between a homogeneous and a non-homogeneous differential equation?

A homogeneous differential equation is one where all terms are a function of the dependent variable and its derivatives, and it equals zero. A non-homogeneous differential equation includes an additional function that is not solely dependent on the variable and its derivatives. Understanding this distinction is important in Pure Mathematics 3 as it determines the method of solution.

Can you explain the concept of a particular solution and a general solution in differential equations?

In differential equations, the general solution includes all possible solutions and typically contains arbitrary constants. A particular solution is a specific instance of the general solution that satisfies initial conditions or specific criteria. In Pure Mathematics 3, finding both types of solutions is key to fully understanding the behavior of differential equations.

Why is it important to understand the initial conditions when solving differential equations in Pure Mathematics 3?

Initial conditions are crucial because they allow you to find the particular solution of a differential equation. They provide specific values for the function or its derivatives at a certain point, which helps in determining the constants in the general solution. This is an important aspect of solving differential equations in the Cambridge International A Levels curriculum, as it ensures the solution is applicable to real-world scenarios.

Why is "Forgetting $+c$ after integration" a common mistake in Differential equations?

Why it happens: Habit. How to avoid it: Always add +c. Apply initial condition AFTER integrating both sides. Source: 9709 Examiner Reports 2022-2024

Why is "Wrong handling when exponentiating $\ln|y| = \ldots + c$" a common mistake in Differential equations?

Why it happens: Treating e^c incorrectly. How to avoid it: y = e^c e^kt = Ae^kt where A absorbs sign and e^c. Just use constant A. Source: 9709 Examiner Reports 2022-2024

Pure Mathematics 3Cambridge International A Levels Mathematics (9709)

Differential equations

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Differential Equations Study Notes — Cambridge International A Level Mathematics 9709 P3 (2024-2026 syllabus)

First-order separable equations. Modelling growth, decay, cooling. Initial conditions to find constants.

What you’ll learn

Mapped to the Cambridge IGCSE 9709 syllabus (2024-2026).

P3.8.1 — Solve separable first-order DEs.
P3.8.2 — Apply initial conditions.
P3.8.3 — Model growth, decay, cooling.

Separable DEs

Variables to opposite sides.

Form. $\frac{dy}{dx} = f(x) g(y)$ . Can be rearranged to put all $y$ on one side, all $x$ on other.

Method.

Multiply or divide to separate: $\frac{dy}{g(y)} = f(x) \, dx$ .
Integrate both sides.
Apply initial condition to find $c$ .
(Optional) solve for $y$ explicitly.

Example. $\frac{dy}{dx} = \frac{x}{y}$ with $y(2) = 3$ .

Separate: $y \, dy = x \, dx$ .
Integrate: $\frac{y^2}{2} = \frac{x^2}{2} + c$ .
IC: $\frac{9}{2} = \frac{4}{2} + c \Rightarrow c = \frac{5}{2}$ .
$y^2 = x^2 + 5$ .

Cambridge tip. Show separation explicitly — mark scheme awards a method mark.

Separate variables.
Integrate both sides.
Apply initial condition.

See the full worked example for differential equations →

Exponential models

$\frac{dP}{dt} = kP$ .

Pattern. Rate of change proportional to current value.

General solution. $P = P_0 e^{kt}$ where:

$k > 0$ → growth.
$k < 0$ → decay.

Derivation. From $\frac{dP}{P} = k \, dt$ . Integrate: $\ln|P| = kt + c$ . Exponentiate: $P = Ae^{kt}$ where $A = e^c$ .

Half-life. For decay $P = P_0 e^{-kt}$ with $k > 0$ : $t_{1/2} = \frac{\ln 2}{k}$ .

Cambridge tip. State exponential solution if you recognise the form — saves time.

$\frac{dP}{dt} = kP$ → $P_0 e^{kt}$ .
Half-life $= \ln 2 / k$ for decay.
Recognise pattern.

See the full worked example for differential equations →

Newton's law of cooling

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

Setup. Object at temperature $T$ in surroundings at $T_a$ . Rate of cooling proportional to temperature DIFFERENCE.

DE. $\frac{dT}{dt} = -k(T - T_a)$ where $k > 0$ .

Solution. Separate: $\frac{dT}{T - T_a} = -k \, dt$ . Integrate: $\ln|T - T_a| = -kt + c$ . Exponentiate: $T - T_a = Ae^{-kt}$ . So $T = T_a + Ae^{-kt}$ .

Interpretation. $T \to T_a$ as $t \to \infty$ . Equilibrium with surroundings.

Example. Object cooling with $T(0) = 80$ °C, $T_a = 20$ °C, $T(5) = 60$ °C.

General: $T = 20 + Ae^{-kt}$ .
IC: $A = 60$ .
$T(5) = 60$ : $60 = 20 + 60e^{-5k} \Rightarrow e^{-5k} = \frac{2}{3} \Rightarrow k = \frac{\ln(3/2)}{5}$ .

Cambridge tip. Identify ambient temperature $T_a$ from the DE coefficient structure.

$\frac{dT}{dt} = -k(T - T_a)$ .
Solution: $T = T_a + Ae^{-kt}$ .
Approaches ambient.

See the full worked example for differential equations →

How it’s examined

DEs appear every P3 — typically 10-12 mark question. Most-tested: separable + IC (8-10 marks), modelling problem (10-12 marks).

Sources: Cambridge International A Level Mathematics 9709 syllabus (2024-2026); 9709 Examiner Reports 2022-2024; 9709/32 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Differential equations, ready to print or save as PDF.

Step-by-step worked examples — Differential equations

Step-by-step solutions to past-paper-style questions on differential equations, written exactly the way a tutor would explain them at the board.

1Separable DE (8 marks)
Extended• Adapted from 9709/32 May/Jun 2024• separable
Question
Solve $\frac{dy}{dx} = \frac{x}{y}$ given $y = 3$ when $x = 2$ . (8 marks)
Step-by-step solution
1. Step 1
  Separate variables. Multiply both sides by $y$ and $dx$ :
  $y \, dy = x \, dx$
2. Step 2
  Integrate both sides.
  $\dfrac{y^2}{2} = \dfrac{x^2}{2} + c$
3. Step 3
  Apply initial condition. $\frac{9}{2} = \frac{4}{2} + c \Rightarrow c = \frac{5}{2}$ .
4. Step 4
  Final.
  $y^2 = x^2 + 5$
Answer
$y^2 = x^2 + 5$ (or $y = \sqrt{x^2 + 5}$ choosing positive root).
2Exponential growth/decay DE (8 marks)
Extended• exponential
Question
A radioactive substance decays so that $\frac{dN}{dt} = -kN$ with $N(0) = N_0$ . Solve. (8 marks)
Step-by-step solution
1. Step 1
  Separate.
  $\dfrac{dN}{N} = -k \, dt$
2. Step 2
  Integrate.
  $\ln|N| = -kt + c$
3. Step 3
  Exponentiate. $N = Ae^{-kt}$ where $A = e^c$ .
4. Step 4
  Apply $N(0) = N_0$ . $N_0 = A \Rightarrow A = N_0$ .
5. Step 5
  Final.
  $N = N_0 e^{-kt}$
Answer
$N = N_0 e^{-kt}$ .
3Newton's law of cooling (10 marks)
Extended• Newton cooling
Question
An object cools according to $\frac{dT}{dt} = -k(T - 20)$ where $T$ is in °C and $t$ in minutes. If $T = 80$ at $t = 0$ and $T = 60$ at $t = 5$ , find $T$ as function of $t$ . (10 marks)
Step-by-step solution
1. Step 1
  Separate variables.
  $\dfrac{dT}{T - 20} = -k \, dt$
2. Step 2
  Integrate.
  $\ln|T - 20| = -kt + c$
3. Step 3
  Exponentiate. $T - 20 = Ae^{-kt}$ .
4. Step 4
  $t = 0$ : $T = 80$ . $80 - 20 = A \Rightarrow A = 60$ .
5. Step 5
  $t = 5$ : $T = 60$ . $60 - 20 = 60 e^{-5k} \Rightarrow e^{-5k} = \frac{2}{3} \Rightarrow k = \frac{1}{5}\ln\frac{3}{2}$ .
6. Step 6
  Final.
  $T = 20 + 60 e^{-kt}, \quad k = \dfrac{1}{5}\ln\dfrac{3}{2}$
Answer
$T = 20 + 60 e^{-kt}$ where $k = \frac{1}{5}\ln\frac{3}{2}$ .

Key Formulae — Differential equations

The formulae you need to memorise for differential equations on the Cambridge International A Level 9709 paper, with every variable defined in plain English and a note on when to use it.

Separable DE
$\dfrac{dy}{dx} = f(x) g(y) \Rightarrow \int \dfrac{dy}{g(y)} = \int f(x) \, dx$
When to use
When DE can be written as $\frac{dy}{dx} = f(x) g(y)$ . Separate variables to opposite sides.
Exponential growth/decay
$\dfrac{dP}{dt} = kP \Rightarrow P = P_0 e^{kt}$
$P_0$
initial value at $t = 0$
When to use
Quantity grows/decays at rate proportional to itself.

Key Definitions and Keywords — Differential equations

Definitions to memorise and the exact keywords mark schemes credit for differential equations answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9709 sitting.

Differential equation
Examiner keyword
Equation involving derivatives. Solution is a function (not a number).
Separable equation
Examiner keyword
First-order DE of form $\frac{dy}{dx} = f(x) g(y)$ . Variables can be separated to opposite sides.
General solution
Solution with arbitrary constant. Family of solutions.
Particular solution
Examiner keyword
Specific solution after using initial condition to determine constant.

Common Mistakes and Misconceptions — Differential equations

The traps other students keep falling into on differential equations questions — taken from recent Cambridge International A Level 9709 examiner reports and mark schemes — and how to avoid them.

✕Forgetting $+c$ after integration
9709 Examiner Reports 2022-2024
Why it happens
Habit.
How to avoid it
Always add $+c$ . Apply initial condition AFTER integrating both sides.
✕Wrong handling when exponentiating $\ln|y| = \ldots + c$
9709 Examiner Reports 2022-2024
Why it happens
Treating $e^c$ incorrectly.
How to avoid it
$y = \pm e^c e^{kt} = Ae^{kt}$ where $A$ absorbs sign and $e^c$ . Just use constant $A$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Differential equations — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Pure Mathematics 3Cambridge International A Levels Mathematics (9709)

Differential equations

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

Previous Next

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Page 1 / 0

Detailed Study Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Differential equations, ready to print or save offline.

Differential Equations Study Notes — Cambridge International A Level Mathematics 9709 P3 (2024-2026 syllabus)

First-order separable equations. Modelling growth, decay, cooling. Initial conditions to find constants.

What you’ll learn

Mapped to the Cambridge IGCSE 9709 syllabus (2024-2026).

P3.8.1 — Solve separable first-order DEs.
P3.8.2 — Apply initial conditions.
P3.8.3 — Model growth, decay, cooling.

Separable DEs

Variables to opposite sides.

Form. $\frac{dy}{dx} = f(x) g(y)$ . Can be rearranged to put all $y$ on one side, all $x$ on other.

Method.

Multiply or divide to separate: $\frac{dy}{g(y)} = f(x) \, dx$ .
Integrate both sides.
Apply initial condition to find $c$ .
(Optional) solve for $y$ explicitly.

Example. $\frac{dy}{dx} = \frac{x}{y}$ with $y(2) = 3$ .

Separate: $y \, dy = x \, dx$ .
Integrate: $\frac{y^2}{2} = \frac{x^2}{2} + c$ .
IC: $\frac{9}{2} = \frac{4}{2} + c \Rightarrow c = \frac{5}{2}$ .
$y^2 = x^2 + 5$ .

Cambridge tip. Show separation explicitly — mark scheme awards a method mark.

Separate variables.
Integrate both sides.
Apply initial condition.

See the full worked example for differential equations →

Exponential models

$\frac{dP}{dt} = kP$ .

Pattern. Rate of change proportional to current value.

General solution. $P = P_0 e^{kt}$ where:

$k > 0$ → growth.
$k < 0$ → decay.

Derivation. From $\frac{dP}{P} = k \, dt$ . Integrate: $\ln|P| = kt + c$ . Exponentiate: $P = Ae^{kt}$ where $A = e^c$ .

Half-life. For decay $P = P_0 e^{-kt}$ with $k > 0$ : $t_{1/2} = \frac{\ln 2}{k}$ .

Cambridge tip. State exponential solution if you recognise the form — saves time.

$\frac{dP}{dt} = kP$ → $P_0 e^{kt}$ .
Half-life $= \ln 2 / k$ for decay.
Recognise pattern.

See the full worked example for differential equations →

Newton's law of cooling

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

Setup. Object at temperature $T$ in surroundings at $T_a$ . Rate of cooling proportional to temperature DIFFERENCE.

DE. $\frac{dT}{dt} = -k(T - T_a)$ where $k > 0$ .

Solution. Separate: $\frac{dT}{T - T_a} = -k \, dt$ . Integrate: $\ln|T - T_a| = -kt + c$ . Exponentiate: $T - T_a = Ae^{-kt}$ . So $T = T_a + Ae^{-kt}$ .

Interpretation. $T \to T_a$ as $t \to \infty$ . Equilibrium with surroundings.

Example. Object cooling with $T(0) = 80$ °C, $T_a = 20$ °C, $T(5) = 60$ °C.

General: $T = 20 + Ae^{-kt}$ .
IC: $A = 60$ .
$T(5) = 60$ : $60 = 20 + 60e^{-5k} \Rightarrow e^{-5k} = \frac{2}{3} \Rightarrow k = \frac{\ln(3/2)}{5}$ .

Cambridge tip. Identify ambient temperature $T_a$ from the DE coefficient structure.

$\frac{dT}{dt} = -k(T - T_a)$ .
Solution: $T = T_a + Ae^{-kt}$ .
Approaches ambient.

See the full worked example for differential equations →

How it’s examined

DEs appear every P3 — typically 10-12 mark question. Most-tested: separable + IC (8-10 marks), modelling problem (10-12 marks).

Sources: Cambridge International A Level Mathematics 9709 syllabus (2024-2026); 9709 Examiner Reports 2022-2024; 9709/32 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Differential equations, ready to print or save as PDF.

Step-by-step worked examples — Differential equations

Step-by-step solutions to past-paper-style questions on differential equations, written exactly the way a tutor would explain them at the board.

1Separable DE (8 marks)
Extended• Adapted from 9709/32 May/Jun 2024• separable
Question
Solve $\frac{dy}{dx} = \frac{x}{y}$ given $y = 3$ when $x = 2$ . (8 marks)
Step-by-step solution
1. Step 1
  Separate variables. Multiply both sides by $y$ and $dx$ :
  $y \, dy = x \, dx$
2. Step 2
  Integrate both sides.
  $\dfrac{y^2}{2} = \dfrac{x^2}{2} + c$
3. Step 3
  Apply initial condition. $\frac{9}{2} = \frac{4}{2} + c \Rightarrow c = \frac{5}{2}$ .
4. Step 4
  Final.
  $y^2 = x^2 + 5$
Answer
$y^2 = x^2 + 5$ (or $y = \sqrt{x^2 + 5}$ choosing positive root).
2Exponential growth/decay DE (8 marks)
Extended• exponential
Question
A radioactive substance decays so that $\frac{dN}{dt} = -kN$ with $N(0) = N_0$ . Solve. (8 marks)
Step-by-step solution
1. Step 1
  Separate.
  $\dfrac{dN}{N} = -k \, dt$
2. Step 2
  Integrate.
  $\ln|N| = -kt + c$
3. Step 3
  Exponentiate. $N = Ae^{-kt}$ where $A = e^c$ .
4. Step 4
  Apply $N(0) = N_0$ . $N_0 = A \Rightarrow A = N_0$ .
5. Step 5
  Final.
  $N = N_0 e^{-kt}$
Answer
$N = N_0 e^{-kt}$ .
3Newton's law of cooling (10 marks)
Extended• Newton cooling
Question
An object cools according to $\frac{dT}{dt} = -k(T - 20)$ where $T$ is in °C and $t$ in minutes. If $T = 80$ at $t = 0$ and $T = 60$ at $t = 5$ , find $T$ as function of $t$ . (10 marks)
Step-by-step solution
1. Step 1
  Separate variables.
  $\dfrac{dT}{T - 20} = -k \, dt$
2. Step 2
  Integrate.
  $\ln|T - 20| = -kt + c$
3. Step 3
  Exponentiate. $T - 20 = Ae^{-kt}$ .
4. Step 4
  $t = 0$ : $T = 80$ . $80 - 20 = A \Rightarrow A = 60$ .
5. Step 5
  $t = 5$ : $T = 60$ . $60 - 20 = 60 e^{-5k} \Rightarrow e^{-5k} = \frac{2}{3} \Rightarrow k = \frac{1}{5}\ln\frac{3}{2}$ .
6. Step 6
  Final.
  $T = 20 + 60 e^{-kt}, \quad k = \dfrac{1}{5}\ln\dfrac{3}{2}$
Answer
$T = 20 + 60 e^{-kt}$ where $k = \frac{1}{5}\ln\frac{3}{2}$ .

Key Formulae — Differential equations

The formulae you need to memorise for differential equations on the Cambridge International A Level 9709 paper, with every variable defined in plain English and a note on when to use it.

Separable DE
$\dfrac{dy}{dx} = f(x) g(y) \Rightarrow \int \dfrac{dy}{g(y)} = \int f(x) \, dx$
When to use
When DE can be written as $\frac{dy}{dx} = f(x) g(y)$ . Separate variables to opposite sides.
Exponential growth/decay
$\dfrac{dP}{dt} = kP \Rightarrow P = P_0 e^{kt}$
$P_0$
initial value at $t = 0$
When to use
Quantity grows/decays at rate proportional to itself.

Key Definitions and Keywords — Differential equations

Differential equation
Examiner keyword
Equation involving derivatives. Solution is a function (not a number).
Separable equation
Examiner keyword
First-order DE of form $\frac{dy}{dx} = f(x) g(y)$ . Variables can be separated to opposite sides.
General solution
Solution with arbitrary constant. Family of solutions.
Particular solution
Examiner keyword
Specific solution after using initial condition to determine constant.

Common Mistakes and Misconceptions — Differential equations

The traps other students keep falling into on differential equations questions — taken from recent Cambridge International A Level 9709 examiner reports and mark schemes — and how to avoid them.

✕Forgetting $+c$ after integration
9709 Examiner Reports 2022-2024
Why it happens
Habit.
How to avoid it
Always add $+c$ . Apply initial condition AFTER integrating both sides.
✕Wrong handling when exponentiating $\ln|y| = \ldots + c$
9709 Examiner Reports 2022-2024
Why it happens
Treating $e^c$ incorrectly.
How to avoid it
$y = \pm e^c e^{kt} = Ae^{kt}$ where $A$ absorbs sign and $e^c$ . Just use constant $A$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Differential equations — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Differential equations

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Separable DE (8 marks)

2Exponential growth/decay DE (8 marks)

3Newton's law of cooling (10 marks)

Separable DE

Exponential growth/decay

Differential equation

Separable equation

General solution

Particular solution

✕Forgetting +c+c+c after integration

✕Wrong handling when exponentiating ln⁡∣y∣=…+c\ln|y| = \ldots + cln∣y∣=…+c

Practice questions

Past paper style quiz

Assess your understanding

Differential equations — frequently asked questions

Differential equations

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Separable DE (8 marks)

2Exponential growth/decay DE (8 marks)

3Newton's law of cooling (10 marks)

Separable DE

Exponential growth/decay

Differential equation

Separable equation

General solution

Particular solution

✕Forgetting +c+c+c after integration

✕Wrong handling when exponentiating ln⁡∣y∣=…+c\ln|y| = \ldots + cln∣y∣=…+c

Practice questions

Past paper style quiz

Assess your understanding

Differential equations — frequently asked questions

✕Forgetting $+c$ after integration

✕Wrong handling when exponentiating $\ln|y| = \ldots + c$

✕Forgetting $+c$ after integration

✕Wrong handling when exponentiating $\ln|y| = \ldots + c$