Notes · Further MathematicsUK · A-Levels

Differential equations

This topic solves the linear differential equations that model change. First-order linear equations are solved with an integrating factor; second-order linear equations with constant coefficients are solved through the auxiliary equation, giving a complementary function and, for a forced equation, a particular integral. The methods are applied to simple harmonic motion, damped and forced oscillations, and coupled systems.

5 sections~13 min reading time3 competenciesLevel Standard 1 · Advanced 4

T·0999 / 13

Exam profile

AO1 · Solve first- and second-order linear differential equations accuratelyAO2 · Derive the general solution via the auxiliary equation and interpret the three casesAO3 · Set up, solve and interpret differential-equation models of oscillating systems

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basic level

AS Further Mathematics covers first-order linear equations by integrating factor and the simplest second-order equations.

higher level

The full A-Level adds the full theory of second-order linear equations, modelling of damped and forced oscillations, and coupled first-order systems.

Depth

Text

Contents · 5 sections

Differential equations

First-order linear equations and integrating factors#

StandardAQA 7367 §3.9DfE Further Mathematics

A first-order linear differential equation can always be written in the standard form

\dfrac{\mathrm{d}y}{\mathrm{d}x} + P(x)\,y = Q(x)

, where

P

and

Q

are functions of

x

alone. The essential first step in every question is to rearrange the equation into exactly this form, dividing through by the coefficient of

\dfrac{\mathrm{d}y}{\mathrm{d}x}

if necessary, since the method depends on identifying

P(x)

correctly.

The equation is solved by multiplying through by the integrating factor

I = \mathrm{e}^{\int P(x)\,\mathrm{d}x}

. The point of this factor is that it turns the left-hand side into the exact derivative of a product: after multiplying, the equation becomes

\dfrac{\mathrm{d}}{\mathrm{d}x}\big(I\,y\big) = I\,Q(x)

. This works because

\dfrac{\mathrm{d}I}{\mathrm{d}x} = P\,I

, so the product rule reconstitutes

I\dfrac{\mathrm{d}y}{\mathrm{d}x} + PIy

as a single derivative.

Once the equation is in the form

\dfrac{\mathrm{d}}{\mathrm{d}x}(Iy) = IQ

, integrating both sides gives

Iy = \displaystyle\int I\,Q(x)\,\mathrm{d}x + c

, and dividing by

I

yields the general solution. The constant of integration

c

is essential — it encodes the family of solution curves — and any given initial condition is applied at the end to fix its value. Omitting

c

, or applying the condition before dividing by

I

, is a frequent slip.

When computing the integrating factor, note that a constant of integration in

\int P\,\mathrm{d}x

is unnecessary (it only multiplies

I

by a constant, which cancels), and that a logarithm in the exponent simplifies:

\mathrm{e}^{\int \frac{n}{x}\,\mathrm{d}x} = \mathrm{e}^{n\ln x} = x^n

. Recognising such simplifications keeps the integrating factor manageable and is often the key to a clean solution.

\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)\,y = Q(x), \qquad I = \mathrm{e}^{\int P(x)\,\mathrm{d}x}, \qquad \frac{\mathrm{d}}{\mathrm{d}x}(Iy) = I\,Q(x)

Integrating-factor method

The factor $I$ makes the left side an exact derivative; integrate and divide by $I$ .

Worked example

Solving by integrating factor

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{2}{x}\,y = x^2$ for $x > 0$ .

Identify P and the integrating factor
$P(x) = \dfrac{2}{x}$ , so $I = \mathrm{e}^{\int \frac{2}{x}\,\mathrm{d}x} = \mathrm{e}^{2\ln x} = x^2$ .
Multiply through
$x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2xy = x^4$ , i.e. $\dfrac{\mathrm{d}}{\mathrm{d}x}(x^2 y) = x^4$ .
$\frac{\mathrm{d}}{\mathrm{d}x}(x^2 y) = x^4$
Integrate
$x^2 y = \dfrac{x^5}{5} + c$ .
Solve for y
$y = \dfrac{x^3}{5} + \dfrac{c}{x^2}$ .

Result: $y = \dfrac{x^3}{5} + \dfrac{c}{x^2}$ .

Typical mistakes

Forgetting to divide through so that the coefficient of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is $1$ before reading off $P(x)$ .
Losing the constant of integration, or applying the initial condition before solving for $y$ .

Active revision

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{2}{x}\,y = x^2$ for $x > 0$ , giving the general solution.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Second-order linear equations: the auxiliary equation#

AdvancedAQA 7367 §3.9DfE Further Mathematics

The three damping behaviours

Fig. 1Complex roots give an under-damped decaying oscillation; a repeated root gives critical damping; distinct real roots give over-damping.

A homogeneous second-order linear equation with constant coefficients has the form

a\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} + b\dfrac{\mathrm{d}y}{\mathrm{d}x} + cy = 0

. Trying a solution

y = \mathrm{e}^{mx}

and dividing by

\mathrm{e}^{mx}

turns the equation into the auxiliary (or characteristic) equation

am^2 + bm + c = 0

, a quadratic in

m

whose roots determine the form of the general solution.

There are three cases, matching the discriminant of the quadratic. If the roots

m_1, m_2

are real and distinct, the general solution is

y = A\mathrm{e}^{m_1 x} + B\mathrm{e}^{m_2 x}

. If the roots are real and equal (

m

repeated), one exponential is not enough and the solution is

y = (A + Bx)\mathrm{e}^{mx}

— the extra factor of

x

supplies the second independent solution. If the roots are complex,

m = p \pm q\mathrm{i}

, the solution is

y = \mathrm{e}^{px}(A\cos qx + B\sin qx)

, an oscillation whose amplitude is modulated by

\mathrm{e}^{px}

The two arbitrary constants

A

and

B

reflect that a second-order equation needs two conditions to pin down a unique solution — typically two initial values, or a value and a derivative. Their presence is not optional: a general solution of a second-order equation always contains exactly two arbitrary constants, and losing one signals an error.

The three cases have a direct physical meaning, previewed in the modelling section: applied to a mass on a spring, distinct real roots give over-damping (a sluggish return with no oscillation), a repeated root gives critical damping (the fastest non-oscillatory return), and complex roots give under-damping (a decaying oscillation). The figure comparing the three behaviours makes the link between the algebra of the discriminant and the qualitative motion vivid, and it is a favourite theme of examiners.

am^2 + bm + c = 0

Auxiliary equation

Obtained by trying $y = \mathrm{e}^{mx}$ ; its roots fix the solution form.

\text{distinct: } A\mathrm{e}^{m_1 x} + B\mathrm{e}^{m_2 x}, \quad \text{repeated: } (A+Bx)\mathrm{e}^{mx}, \quad \text{complex } p\pm q\mathrm{i}: \mathrm{e}^{px}(A\cos qx + B\sin qx)

The three cases

The nature of the auxiliary roots determines the general solution.

Worked example

Distinct real roots

Find the general solution of $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 5\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6y = 0$ .

Form the auxiliary equation
Trying $y = \mathrm{e}^{mx}$ gives $m^2 - 5m + 6 = 0$ .
Solve for m
$(m-2)(m-3) = 0$ , so $m = 2$ or $m = 3$ — real and distinct.
$m = 2 \ \text{or}\ m = 3$
Write the general solution
$y = A\mathrm{e}^{2x} + B\mathrm{e}^{3x}$ .

Result: $y = A\mathrm{e}^{2x} + B\mathrm{e}^{3x}$ .

Typical mistakes

Using $y = A\mathrm{e}^{mx} + B\mathrm{e}^{mx}$ for a repeated root (this collapses to one constant) instead of $(A+Bx)\mathrm{e}^{mx}$ .
Dropping the exponential envelope $\mathrm{e}^{px}$ when the roots are complex, or confusing $p$ and $q$ .

Active revision

Find the general solution of $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 5\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6y = 0$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Particular integrals and the general solution#

AdvancedAQA 7367 §3.9DfE Further Mathematics

When the right-hand side is not zero —

a y'' + b y' + cy = f(x)

— the general solution is the sum of two parts: the complementary function (CF), which is the general solution of the associated homogeneous equation found via the auxiliary equation, and a particular integral (PI), which is any single solution of the full equation. So

y = \text{CF} + \text{PI}

, and this decomposition is the organising idea of the whole method.

The particular integral is found by trying a function of the same form as

f(x)

with undetermined coefficients, then substituting to fix them. If

f(x)

is a polynomial, try a general polynomial of the same degree; if

f(x) = k\mathrm{e}^{\lambda x}

, try

C\mathrm{e}^{\lambda x}

; if

f(x)

involves

\cos\omega x

\sin\omega x

, try

C\cos\omega x + D\sin\omega x

(both terms, even if only one appears in

f

). Substituting the trial into the equation and equating coefficients determines the constants.

A crucial refinement handles the case where the trial function is already part of the complementary function. If, say,

f(x) = k\mathrm{e}^{\lambda x}

but

\mathrm{e}^{\lambda x}

already appears in the CF (because

\lambda

is an auxiliary root), then

C\mathrm{e}^{\lambda x}

substitutes to give zero and cannot work; the remedy is to multiply the trial by

x

(or

x^2

for a repeated root). Spotting this clash before substituting saves a wasted attempt.

Once the PI is found, the general solution

y = \text{CF} + \text{PI}

contains the two arbitrary constants from the CF, and these are fixed last using the boundary or initial conditions. It is essential to apply the conditions to the complete general solution, not to the CF alone — a common and costly error — because the PI contributes to the value and derivative of

y

y = \underbrace{\text{CF}}_{\text{homogeneous solution}} + \underbrace{\text{PI}}_{\text{one solution of the full equation}}

Structure of the general solution

Complementary function plus a particular integral; the two CF constants are fixed by the conditions.

Worked example

Complementary function plus particular integral

Find the general solution of $y'' - 3y' + 2y = \mathrm{e}^{3x}$ .

Complementary function
Auxiliary: $m^2 - 3m + 2 = (m-1)(m-2) = 0$ , so $m = 1, 2$ and $\text{CF} = A\mathrm{e}^{x} + B\mathrm{e}^{2x}$ .
Choose a trial PI
Since $\mathrm{e}^{3x}$ is not in the CF, try $y_p = C\mathrm{e}^{3x}$ ; then $y_p'' = 9C\mathrm{e}^{3x}$ , $y_p' = 3C\mathrm{e}^{3x}$ .
Substitute and equate
$9C - 3(3C) + 2C = 1$ , i.e. $(9 - 9 + 2)C = 1$ , so $2C = 1$ and $C = \tfrac12$ .
$y_p = \tfrac{1}{2}\mathrm{e}^{3x}$
Combine
$y = A\mathrm{e}^{x} + B\mathrm{e}^{2x} + \tfrac{1}{2}\mathrm{e}^{3x}$ .

Result: $y = A\mathrm{e}^{x} + B\mathrm{e}^{2x} + \tfrac{1}{2}\mathrm{e}^{3x}$ .

Typical mistakes

Trying only $\cos\omega x$ (or only $\sin\omega x$ ) when $f(x)$ is trigonometric — both terms are needed.
Applying initial conditions to the complementary function before adding the particular integral.

Active revision

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

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Modelling oscillations: SHM, damping and forcing#

AdvancedAQA 7367 §3.9DfE Further Mathematics

Simple harmonic motion

Fig. 2The undamped solution

x = 2\cos(2t)

oscillates with constant amplitude

2

and period

\pi

Simple harmonic motion is the model

\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} = -\omega^2 x

, whose auxiliary equation

m^2 + \omega^2 = 0

has purely imaginary roots

m = \pm\omega\mathrm{i}

. The general solution is

x = A\cos\omega t + B\sin\omega t

, an undamped oscillation of angular frequency

\omega

and period

\dfrac{2\pi}{\omega}

. It can be rewritten as

x = R\cos(\omega t - \varepsilon)

, displaying the amplitude

R = \sqrt{A^2 + B^2}

and the phase

\varepsilon

directly.

Adding resistance proportional to velocity gives the damped equation

m\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} + c\dfrac{\mathrm{d}x}{\mathrm{d}t} + kx = 0

. The discriminant of its auxiliary equation decides the behaviour: complex roots (light damping) give an under-damped decaying oscillation

x = \mathrm{e}^{-\beta t}(A\cos\omega t + B\sin\omega t)

, whose amplitude decays inside the envelope

\pm\mathrm{e}^{-\beta t}

; a repeated root gives critical damping (the quickest return with no oscillation); and distinct real roots give over-damping (a slow, sluggish return). These are exactly the three cases of the previous sections seen in a physical setting.

Applying a periodic driving force gives a forced (inhomogeneous) equation such as

m\ddot x + c\dot x + kx = F_0\cos\Omega t

. The complementary function represents the transient — the system's own damped motion, which dies away — while the particular integral is the steady state, an oscillation at the driving frequency

\Omega

. The long-term behaviour is therefore governed entirely by the PI, and resonance occurs when the driving frequency is close to the system's natural frequency, producing a large steady-state amplitude.

Setting up these models correctly is a modelling (AO3) skill: read the physical description, write down the equation with the right signs (restoring force negative, resistance opposing velocity), and interpret the solution back in the physical context — the period of an oscillation, the time for an amplitude to halve, whether the system oscillates at all. The figures of

x = A\cos\omega t

and of a decaying oscillation within its envelope connect the algebraic solution to the motion it describes.

\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} = -\omega^2 x \ \Rightarrow\ x = A\cos\omega t + B\sin\omega t = R\cos(\omega t - \varepsilon), \quad T = \frac{2\pi}{\omega}

Simple harmonic motion

Undamped oscillation of angular frequency $\omega$ ; amplitude $R = \sqrt{A^2+B^2}$ .

m\ddot x + c\dot x + kx = 0 \ \Rightarrow\ x = \mathrm{e}^{-\beta t}(A\cos\omega t + B\sin\omega t) \ (\text{under-damped})

Damped oscillation

The amplitude decays inside the envelope $\pm\mathrm{e}^{-\beta t}$ .

An under-damped oscillation and its envelope

Fig. 3The solution

x = \mathrm{e}^{-0.3t}\cos(3t)

oscillates within the decaying envelope

\pm\mathrm{e}^{-0.3t}

Worked example

Simple harmonic motion from initial conditions

Solve $\ddot x + 9x = 0$ with $x(0) = 2$ , $\dot x(0) = 0$ , and state the period.

Identify SHM
Comparing with $\ddot x = -\omega^2 x$ gives $\omega^2 = 9$ , so $\omega = 3$ and $x = A\cos 3t + B\sin 3t$ .
Apply x(0) = 2
At $t = 0$ : $x = A = 2$ .
Apply the velocity condition
$\dot x = -3A\sin 3t + 3B\cos 3t$ ; at $t = 0$ , $\dot x = 3B = 0$ , so $B = 0$ .
$x = 2\cos 3t$
Period
$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{3}$ .

Result: $x = 2\cos 3t$ , with period $\dfrac{2\pi}{3}$ .

Typical mistakes

Getting the sign of the restoring term wrong, turning oscillation into exponential growth.
Confusing the transient and the steady state, or forgetting that the CF dies away in a damped forced system.

Active revision

A particle moves so that $\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 9x = 0$ , with $x = 2$ and $\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0$ when $t = 0$ . Find $x$ as a function of $t$ and state the period.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Coupled first-order systems#

AdvancedAQA 7367 §3.9DfE Further Mathematics

A coupled system is a pair of first-order equations in which each rate of change depends on both variables, for example

\dfrac{\mathrm{d}x}{\mathrm{d}t} = ax + by

and

\dfrac{\mathrm{d}y}{\mathrm{d}t} = cx + dy

. Such systems model interacting quantities — two connected tanks, two coupled circuits, predator and prey in a linearised model — where neither variable evolves independently of the other.

The standard A-Level method reduces the coupled pair to a single second-order equation in one variable. Differentiate one equation with respect to

t

, then use the two original equations to eliminate the other variable and its derivative. This produces a second-order linear equation in, say,

x

alone, which is solved by the auxiliary-equation method of the earlier sections.

Having solved for

x

, the second variable

y

is recovered by substituting

x

back into one of the original first-order equations and rearranging — no new integration is needed, so

y

inherits the same arbitrary constants as

x

rather than introducing fresh ones. This is an important consistency check: a correctly solved

2\times 2

system has exactly two arbitrary constants in total, shared between

x

and

y

Initial conditions for both variables are then applied to the pair of solutions to fix the two constants. The elimination method keeps the whole problem within techniques already mastered — differentiation, substitution and the auxiliary equation — and its main pitfall is algebraic: care is needed to eliminate cleanly and to recover

y

without accidentally introducing a spurious extra constant.

\frac{\mathrm{d}x}{\mathrm{d}t} = ax + by, \qquad \frac{\mathrm{d}y}{\mathrm{d}t} = cx + dy

A coupled linear system

Reduce to one second-order equation by differentiation and elimination.

Worked example

Reducing a coupled system

Solve $\dfrac{\mathrm{d}x}{\mathrm{d}t} = y$ , $\dfrac{\mathrm{d}y}{\mathrm{d}t} = -x$ .

Differentiate the first equation
$\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} = \dfrac{\mathrm{d}y}{\mathrm{d}t}$ .
Eliminate y
Substitute $\dfrac{\mathrm{d}y}{\mathrm{d}t} = -x$ : $\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} = -x$ , i.e. $\ddot x + x = 0$ .
$\ddot x + x = 0$
Solve for x
This is SHM with $\omega = 1$ : $x = A\cos t + B\sin t$ .
Recover y
Since $y = \dfrac{\mathrm{d}x}{\mathrm{d}t} = -A\sin t + B\cos t$ . The point $(x,y)$ moves on a circle of radius $\sqrt{A^2+B^2}$ .

Result: $x = A\cos t + B\sin t$ , $y = -A\sin t + B\cos t$ ; the motion is circular in the $(x,y)$ plane.

Typical mistakes

Introducing a new constant of integration when recovering the second variable, giving three constants for a $2\times 2$ system.
Algebraic slips in the elimination, especially sign errors when substituting the original equations.

Active revision

Solve the system $\dfrac{\mathrm{d}x}{\mathrm{d}t} = y$ , $\dfrac{\mathrm{d}y}{\mathrm{d}t} = -x$ , giving $x$ and $y$ as functions of $t$ , and describe the motion in the $(x,y)$ plane.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Differential equations

Key points

Solving by integrating factor

Key points

Distinct real roots

Key points

Complementary function plus particular integral

Key points

Simple harmonic motion from initial conditions

Key points

Reducing a coupled system

Sources