Differential Equations

CIE A-Level Further Mathematics· Unit 2: Further Pure 2· 15 min read

1. First-Order Linear Differential Equations★★☆☆☆⏱ 10 min

📘 Definition

First-order linear differential equation

\frac{dy}{dx} + P(x)y = Q(x)

A first-order equation where $y$ and its derivative are both to the first power, with coefficients that can be functions of the independent variable $x$ .

Example:

\frac{dy}{dx} + 3y = e^x

To solve an equation in standard form above, we first calculate the integrating factor $I (x) = e^{\int P (x) d x}$ . Multiply both sides of the equation by $I (x)$ : the left-hand side simplifies to $\frac{d}{d x} (I (x) y)$ by the product rule. Integrate both sides to solve for $y$ .

📐 Worked Example

Solve $\frac{d y}{d x} + \frac{y}{x} = x^{2}$ for $x > 0$ , given $y = 2$ when $x = 1$ .

1
1. Identify $P (x) = \frac{1}{x}$ and $Q (x) = x^{2}$ from standard form.
2
1. Calculate the integrating factor:
3
$\int P(x) dx = \int \frac{1}{x} dx = \ln x \implies I(x) = e^{\ln x} = x$
4
1. Multiply both sides of the equation by $I (x)$ :
5
$x \frac{dy}{dx} + y = x^3$
6
1. Rewrite left-hand side as product rule derivative:
7
$\frac{d}{dx}\left(xy\right) = x^3$
8
1. Integrate both sides, adding constant of integration:
9
$xy = \frac{x^4}{4} + C$
10
1. Apply boundary condition $y = 2, x = 1$ to find $C$ :
11
$(1)(2) = \frac{1^4}{4} + C \implies C = 2 - \frac{1}{4} = \frac{7}{4}$
12
1. Rearrange to get the final solution:
13
$y = \frac{x^3}{4} + \frac{7}{4x}$

Exam tip:

Always rearrange your equation to get a coefficient of 1 for $\frac{d y}{d x}$ before calculating the integrating factor, to avoid errors in the sign or value of $P (x)$ .

2. Homogeneous Second-Order Differential Equations★★★☆☆⏱ 12 min

📘 Definition

Homogeneous second-order linear DE (constant coefficients)

a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = 0 \quad (a \neq 0)

Second-order equation where all terms involve $y$ or its derivatives, with constant coefficients $a, b, c$ and zero right-hand side.

We solve these by forming the auxiliary (characteristic) quadratic equation $a m^{2} + bm + c = 0$ . The form of the general solution depends entirely on the nature of the roots of this quadratic:

Two distinct real roots $m_{1}, m_{2}$ : $y = A e^{m_{1} x} + B e^{m_{2} x}$
One repeated real root $m$ : $y = (A + B x) e^{m x}$
Two complex conjugate roots $α \pm i β$ : $y = e^{α x} (A cos β x + B sin β x)$

📐 Worked Example

Find the general solution of $\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 0$ .

1
1. Form the auxiliary equation from the coefficients:
2
$m^2 - 4m + 4 = 0$
3
1. Factorize to find roots:
4
$(m-2)^2 = 0 \implies m = 2 \text{ (repeated root)}$
5
1. Use the general solution formula for repeated roots:
6
$y = (A + Bx) e^{2x}$

3. Non-Homogeneous Second-Order Differential Equations★★★★☆⏱ 15 min

📘 Definition

Non-homogeneous second-order linear DE (constant coefficients)

a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = f(x) \quad (a \neq 0, f(x) \neq 0)

General solution = Complementary Function (CF, solution of the homogeneous equation) + Particular Integral (PI, any solution that satisfies the full non-homogeneous equation).

We find the particular integral by the method of undetermined coefficients: guess a form of PI based on $f (x)$ , substitute into the equation to solve for unknown coefficients. If your guess matches a term in the complementary function, multiply it by $x$ (or $x^{2}$ for a repeated root) to get a linearly independent solution.

📐 Worked Example

Find the general solution of $\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 6 e^{4 x}$ .

1
1. First find the complementary function by solving the homogeneous equation:
2
$m^2 - 3m + 2 = 0 \implies (m-1)(m-2) = 0 \implies m = 1, 2$
3
So the complementary function is:
4
$CF: y = A e^x + B e^{2x}$
5
1. Guess the particular integral: $f (x) = 6 e^{4 x}$ , and 4 is not a root of the auxiliary equation, so try $y = C e^{4 x}$ .
6
1. Calculate derivatives and substitute into the left-hand side:
7
$\frac{dy}{dx} = 4C e^{4x}, \quad \frac{d^2y}{dx^2} = 16C e^{4x}$
8
$16C e^{4x} - 3(4C e^{4x}) + 2(C e^{4x}) = 6C e^{4x}$
9
1. Equate to right-hand side to solve for C:
10
$6C e^{4x} = 6 e^{4x} \implies C = 1$
11
1. The general solution is $y = C F + P I$ :
12
$y = A e^x + B e^{2x} + e^{4x}$

4. Common Pitfalls

Wrong move:

Calculating the integrating factor without rearranging first-order DE to standard form.

Why:

If the coefficient of $\frac{d y}{d x}$ is not 1, $P (x)$ will be wrong leading to an incorrect solution.

Correct move:

Always divide through by the coefficient of $\frac{d y}{d x}$ to get it equal to 1 before calculating $I (x)$ .

Wrong move:

Writing $y = A e^{m x} + B e^{m x} = (A + B) e^{m x}$ for repeated root auxiliary equation.

Why:

This reduces to only one arbitrary constant, which is incorrect for a second-order equation.

Correct move:

For repeated root $m$ , use $y = (A + B x) e^{m x}$ which has two independent arbitrary constants.

Wrong move:

Using the same PI guess even when $f (x)$ matches the complementary function.

Why:

Substituting a PI that is a solution of the homogeneous equation will give 0 on the left-hand side, which can never equal $f (x) \neq = 0$ .

Correct move:

Multiply your PI guess by $x$ (or $x^{2}$ for a repeated root) to get a linearly independent solution, then substitute.

Wrong move:

Forgetting to add the constant of integration after integrating first-order DE.

Why:

Leaving out $C$ means you lose the arbitrary constant required for a general solution, and cannot find the correct particular solution for boundary conditions.

Correct move:

Add $C$ immediately after integrating, before applying any boundary conditions to solve for its value.

5. Quick Reference Cheatsheet

Equation Type	General Solution
First-order linear: $\frac{d y}{d x} + P (x) y = Q (x)$	$I = e^{\int P d x}, y = \frac{1}{I} (\int I Q d x + C)$
Homogeneous 2nd order: distinct real roots $m_{1} \neq = m_{2}$	$y = A e^{m_{1} x} + B e^{m_{2} x}$
Homogeneous 2nd order: repeated root $m$	$y = (A + B x) e^{m x}$
Homogeneous 2nd order: complex roots $α \pm i β$	$y = e^{α x} (A cos β x + B sin β x)$
Non-homogeneous 2nd order	$y = Complementary Function + Particular Integral$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Second-order non-homogeneous solution
2023 · 2
First-order integrating factor problem
2024 · 1
Boundary value second-order problem

Going deeper

syllabusCIE 9231 Further Mathematics Syllabus

What's Next

Differential equations are one of the most heavily examined topics in CIE 9231, appearing in both Further Pure papers and often combined with mechanics modelling questions. Solid mastery of the methods covered here is critical to securing high marks in the exam. This topic forms the foundation for more advanced differential equation topics in further study, and for applied problems across A-Level Further Mathematics. Next, you can explore how to use these methods to model real-world systems, or move on to other core FP2 topics.

→Further Mechanics
→Momentum and Impulse