Differential equations

CIE A-Level Mathematics· 9709 Pure Mathematics 3: Topic 8 Differential equations· 20 min read

1. Forming Differential Equations★★☆☆☆⏱ 7 min

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.

📘 Definition

Differential Equation

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

Example:

$\frac{d y}{d x} = 2 x y$ is a first-order differential equation

📐 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
$\frac{dm}{dt} = -km$

2. Solving Separable First-Order Differential Equations★★★☆☆⏱ 8 min

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.

📘 Definition

Separable Differential Equation

A first-order differential equation that can be rewritten in the form $\frac{d y}{d x} = f (x) g (y)$

Example:

$\frac{d y}{d x} = x y^{2}$ is separable, $\frac{d y}{d x} = x + y$ is not

📐 Worked Example

Find the general solution of $\frac{d y}{d x} = \frac{e ^{x} y}{y + 1}$ .

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
$y + \ln|y| = e^x + C$

3. Finding Particular Solutions★★★☆☆⏱ 5 min

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 $\frac{d y}{d x} = 2 y$ 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
$\ln y = 2x + \ln 4 \implies y = 4e^{2x}$

4. Interpreting Solutions and Long-Run Behaviour★★★★☆⏱ 8 min

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 $\frac{d x}{d t} = k x (a - x)$ , 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 $\frac{d P}{d t} = \frac{1}{2} P (4 - P)$ , 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 $P = \frac{4}{3 e ^{- 2 t} + 1}$ , as $t \to \infty$ the term $e^{- 2 t} \to 0$ , so $P \to 4$ . The population approaches a limiting value of 4 thousand — the carrying capacity of the model.

5. Common Pitfalls

Wrong move:

Forgetting a negative sign for a decreasing rate of change

Why:

Most students remember proportionality, but miss that 'rate of decrease' means the derivative is negative

Correct move:

Always add a negative sign for decreasing quantities: a decreasing mass gives $\frac{d m}{d t} = - k m$ for positive $k$

Wrong move:

Adding a constant of integration to both sides after separating variables

Why:

Two arbitrary constants can be combined into one, so this leads to unnecessary, incorrect simplification

Correct move:

Only add one arbitrary constant, to the side with independent variable terms

Wrong move:

Dropping the absolute value when integrating $\frac{1}{y}$

Why:

This can lead to incorrect signs for $y$ when you exponentiate to simplify

Correct move:

Keep the absolute value until you substitute the initial condition to confirm the sign of $y$

Wrong move:

Not checking the particular solution against the original differential equation

Why:

Integration errors or rearrangement mistakes are common, and this check catches them quickly

Correct move:

Differentiate your solution, substitute back into the original DE, and confirm both sides match

6. Quick Reference Cheatsheet

Step	Action for Separable Differential Equations
1	Rearrange into $\frac{1}{g ( y )} d y = f (x) d x$
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

Going deeper

syllabusCIE 9709 Official Pure 3 SyllabusCheck for latest syllabus updates

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.