# Differential equations

> CIE A-Level Mathematics · Pure Mathematics 3 Unit 2
> Source: https://www.owlsprep.com/study/cie-9709-u2-differential-equations/

This module covers first-order separable differential equations for CIE A-Level 9709 Pure 3, including forming equations from context, finding general solutions, and calculating particular solutions from initial conditions.

**Prerequisites:** [Indefinite integration of basic functions](https://www.owlsprep.com/study/cie-9709-u1-indefinite-integration/); [Basic differentiation rules](https://www.owlsprep.com/study/cie-9709-u1-differentiation/)

## Learning objectives

- Form differential equations from contextual and abstract problems
- Identify and solve first-order separable differential equations
- Find particular solutions given initial conditions
- Interpret a solution in context, including its long-run (limiting) behaviour as $t \to \infty$

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P3 past papers (2016–2025): **Differential equations** appears **63 times** in the last 10 years — **9.7%** of all P3 questions (63 of 651).

Most-tested forms: First-order ODE by separation of variables (58), Setting up and solving DEs in modelling contexts (18).

## Forming Differential Equations

A differential equation relates a function to its derivatives. In CIE 9709, you will often be asked to form a differential equation from a descriptive problem, typically involving rates of change.

**Differential Equation** — An equation containing at least one derivative of a dependent variable with respect to an independent variable

*Example:* $\frac{dy}{dx} = 2xy$ is a first-order differential equation

> **tip**
>
> When forming equations from context, always identify which quantity is changing, what it is proportional to, and the sign of the proportionality constant. For decay, the rate is negative.

**Worked example:** The rate of decrease of the mass $m$ of a radioactive substance is proportional to the mass at time $t$. Form a differential equation for this relationship.

1. First, identify the rate of change: a rate of decrease means the derivative is negative:
2. $$-\frac{dm}{dt}$$
3. This rate is proportional to the current mass $m$, so:
4. $$-\frac{dm}{dt} \propto m$$
5. Add a positive constant of proportionality $k$ to get the final equation:
6. $$\frac{dm}{dt} = -km$$

## Solving Separable First-Order Differential Equations

A first-order differential equation is separable if we can rearrange it to group all $y$ terms on the left-hand side and all $x$ terms on the right-hand side. We then integrate both sides to get the general solution.

**Separable Differential Equation** — A first-order differential equation that can be rewritten in the form $\frac{dy}{dx} = f(x)g(y)$

*Example:* $\frac{dy}{dx} = xy^2$ is separable, $\frac{dy}{dx} = x+y$ is not

**Worked example:** Find the general solution of $\frac{dy}{dx} = \frac{e^x y}{y + 1}$.

1. Rearrange to separate variables:
2. $$\frac{y + 1}{y} dy = e^x dx$$
3. Simplify the left-hand side:
4. $$\left(1 + \frac{1}{y}\right) dy = e^x dx$$
5. Integrate both sides, only add one arbitrary constant:
6. $$\int \left(1 + \frac{1}{y}\right) dy = \int e^x dx$$
7. Evaluate the integrals to get the general solution:
8. $$y + \ln|y| = e^x + C$$

## Finding Particular Solutions

A general solution includes an arbitrary constant of integration. When an initial condition (a matching pair of values for the dependent and independent variable) is given, we substitute these values to find the constant, resulting in a unique particular solution.

**Worked example:** Given $\frac{dy}{dx} = 2y$ and $y=4$ when $x=0$, find the particular solution.

1. Separate variables and integrate to get the general solution:
2. $$\int \frac{1}{y} dy = \int 2 dx \implies \ln|y| = 2x + C$$
3. Substitute the initial condition $y=4, x=0$ to find $C$:
4. $$\ln 4 = 2(0) + C \implies C = \ln 4$$
5. Substitute $C$ back and rearrange for $y$:
6. $$\ln y = 2x + \ln 4 \implies y = 4e^{2x}$$

> **tip**
>
> Always check your solution by substituting back into the original differential equation to confirm it works.

## Interpreting Solutions and Long-Run Behaviour

Once you have a solution, an exam will often ask you to interpret it in context — especially its behaviour as $t \to \infty$. For growth or decay towards a fixed level, the exponential term dies away and the solution approaches a constant limiting value. Finding this limit usually means letting $t \to \infty$ and seeing which terms vanish. Models of restricted (logistic) growth, such as $\frac{dx}{dt} = kx(a - x)$, are integrated using partial fractions — a Pure 3 technique — and approach the limiting value $x = a$.

**Worked example:** A population $P$ (in thousands) grows according to the logistic model $\frac{dP}{dt} = \frac{1}{2} P(4 - P)$, where $t$ is measured in years. Given that $P = 1$ when $t = 0$, find $P$ in terms of $t$ and state the value that $P$ approaches in the long term.

1. Separate the variables — the left-hand side will need partial fractions:
2. $$\frac{1}{P(4 - P)}\, dP = \frac{1}{2}\, dt$$
3. Split the left-hand side into partial fractions:
4. $$\frac{1}{P(4 - P)} = \frac{1}{4}\left(\frac{1}{P} + \frac{1}{4 - P}\right)$$
5. Integrate both sides, adding a single arbitrary constant:
6. $$\frac{1}{4}\left(\ln P - \ln(4 - P)\right) = \frac{1}{2}t + C$$
7. Combine the logarithms and multiply through by 4:
8. $$\ln\!\left(\frac{P}{4 - P}\right) = 2t + C_1$$
9. Apply the initial condition $P = 1,\ t = 0$ to find the constant:
10. $$\ln\tfrac{1}{3} = C_1 \implies C_1 = -\ln 3$$
11. Rearrange to make $P$ the subject:
12. $$\frac{P}{4 - P} = \tfrac{1}{3}e^{2t} \implies P = \frac{4e^{2t}}{3 + e^{2t}}$$
13. Interpret the long-run behaviour. Writing $P = \dfrac{4}{3e^{-2t} + 1}$, as $t \to \infty$ the term $e^{-2t} \to 0$, so $P \to 4$. The population approaches a limiting value of 4 thousand — the carrying capacity of the model.

> **tip**
>
> To find a limiting value, rearrange the solution so the time-dependent part appears as $e^{-\lambda t}$ with $\lambda > 0$, then let $t \to \infty$ so that term tends to $0$. Whatever remains is the limiting value.

## Common pitfalls

- **Wrong:** Forgetting a negative sign for a decreasing rate of change
  - Why it fails: Most students remember proportionality, but miss that 'rate of decrease' means the derivative is negative
  - Correct: Always add a negative sign for decreasing quantities: a decreasing mass gives $\frac{dm}{dt} = -km$ for positive $k$
- **Wrong:** Adding a constant of integration to both sides after separating variables
  - Why it fails: Two arbitrary constants can be combined into one, so this leads to unnecessary, incorrect simplification
  - Correct: Only add one arbitrary constant, to the side with independent variable terms
- **Wrong:** Dropping the absolute value when integrating $\frac{1}{y}$
  - Why it fails: This can lead to incorrect signs for $y$ when you exponentiate to simplify
  - Correct: Keep the absolute value until you substitute the initial condition to confirm the sign of $y$
- **Wrong:** Not checking the particular solution against the original differential equation
  - Why it fails: Integration errors or rearrangement mistakes are common, and this check catches them quickly
  - Correct: Differentiate your solution, substitute back into the original DE, and confirm both sides match

## Cheatsheet

| Step | Action for Separable Differential Equations |
| --- | --- |
| 1 | Rearrange into $\frac{1}{g(y)} dy = f(x) dx$ |
| 2 | Integrate both sides, add one arbitrary constant $C$ |
| 3 | Substitute initial condition to find value of $C$ |
| 4 | Rearrange for $y$ and simplify if required |
| 5 | Verify solution in original differential equation |

## What's next

Differential equations are a core calculus topic with wide applications across physics, chemistry, economics, and engineering. Separating variables is the only differential-equation method in 9709; integrating factors and second-order equations belong to Further Mathematics 9231. To go further within 9709, focus on becoming fluent with the standard Pure 3 integration techniques that these equations rely on, especially partial fractions, since the harder marks come from the integration and the interpretation rather than from any new solution method. Differential equations also appear frequently in mechanics problems involving kinematics, so the basics you learn here will support your work in other units of the syllabus.

- [Complex Numbers](https://www.owlsprep.com/study/cie-9709-u2-complex-numbers/)
- [Mechanics](https://www.owlsprep.com/study/cie-9709-u3-overview/)
- [Kinematics](https://www.owlsprep.com/study/cie-9709-u3-kinematics/)

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