HookThe bridge that swayed for two days — and the equation that fixed it
On 10 June 2000 the Millennium Bridge opened across the Thames, and tens of thousands of Londoners walked onto it. Within minutes the deck began to sway sideways. Pedestrians instinctively widened their stance and fell into step with the sway — which fed energy back in at exactly the frequency the bridge wanted to move at, and the swing grew to around 70 millimetres. Two days later the bridge was closed. Arup's engineers did not stiffen it or rebuild it; they spent roughly £5 million fitting 37 fluid-viscous dampers and dozens of tuned mass dampers, then reopened it in February 2002. It has not wobbled since.
What the engineers changed was one coefficient in one equation. The lateral motion of the deck obeys \(m\ddot{x}+c\dot{x}+kx=f(t)\) — mass times acceleration, plus a resistance term proportional to velocity, plus a spring-like restoring force, driven by the crowd. The bridge as built had \(c\) too small: a lightly damped system that oscillates and lets amplitude grow. The dampers raised \(c\) until the mathematics changed character. That single second-order equation, and the first-order equations that lead up to it, are this entire section: how to solve them, how the coefficients dictate the behaviour, and how to build them from a physical situation in the first place — because AQA will ask you to do all three.
ModelFirst order: the integrating factor
A first-order equation is linear when it can be written \(\frac{dy}{dx}+P(x)y=Q(x)\) — \(y\) and \(\frac{dy}{dx}\) appear to the first power only, never multiplied together, never inside another function. Recognising the form is half the skill: an equation like \(x\frac{dy}{dx}+2y=x^3\) qualifies only after you divide through by \(x\) so the derivative stands alone, while \(\frac{dy}{dx}=y^2\) never qualifies at all, however you rearrange it. Check the form before reaching for the method — the spec explicitly asks you to recognise when the technique applies.
The method multiplies the whole equation by the integrating factor \(\mu=e^{\int P\,dx}\), chosen for one reason: it turns the left side into an exact derivative, \(\frac{d}{dx}(\mu y)=\mu Q\). One integration then gives \(\mu y=\int \mu Q\,dx+c\). The constant \(c\) is the difference between a general solution — a whole family of curves, one for each \(c\) — and a particular solution, the single member forced through a given point by a boundary or initial condition. Exam questions almost always want the particular solution, and the mark scheme wants to see \(c\) appear before the condition is applied, not bolted on at the end.
Solve \(\frac{dy}{dx}+\frac{2}{x}y=x\) given that \(y=1\) when \(x=1\).
Step 1 — integrating factor: \(P(x)=\frac{2}{x}\), so \(\mu=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2\).
Step 2 — multiply and collapse: \(x^2\frac{dy}{dx}+2xy=x^3\), and the left side is exactly \(\frac{d}{dx}(x^2y)\).
Step 3 — integrate: \(x^2y=\int x^3\,dx=\frac{x^4}{4}+c\), so the general solution is \(y=\frac{x^2}{4}+\frac{c}{x^2}\).
Step 4 — apply the condition: \(y(1)=1\) gives \(1=\frac{1}{4}+c\), so \(c=\frac{3}{4}\) and the particular solution is \(y=\frac{x^2}{4}+\frac{3}{4x^2}\).
MechanismSecond order homogeneous: the auxiliary equation and its discriminant
For \(y''+ay'+by=0\), try \(y=e^{mx}\). Substituting gives \(e^{mx}(m^2+am+b)=0\), and since the exponential is never zero, everything hangs on the auxiliary equation \(m^2+am+b=0\). A quadratic has three possible root structures, and the discriminant \(a^2-4b\) tells you which — which is exactly why AQA lists 'link the sign of the discriminant to the form of the solution' as its own skill.
Two distinct real roots (\(a^2-4b>0\)): \(y=Ae^{m_1x}+Be^{m_2x}\) — pure exponential growth or decay, no oscillation. One repeated root (\(a^2-4b=0\)): \(y=(A+Bx)e^{mx}\) — the extra factor of \(x\) is compulsory, because \(Ae^{mx}\) alone is only one independent solution and a second-order equation needs two. Complex conjugate roots \(m=\alpha\pm\beta i\) (\(a^2-4b<0\)): \(y=e^{\alpha x}(A\cos\beta x+B\sin\beta x)\) — genuine oscillation, wrapped in an exponential envelope that grows if \(\alpha>0\) and dies if \(\alpha<0\). Read the physics off the roots before solving anything: real part sets the envelope, imaginary part sets the frequency.
Solve \(y''+4y'+13y=0\) with \(y(0)=2\) and \(y'(0)=1\).
Auxiliary equation: \(m^2+4m+13=0\), so \(m=\frac{-4\pm\sqrt{16-52}}{2}=-2\pm 3i\). The discriminant is negative, so the solution oscillates inside a decaying envelope.
General solution: \(y=e^{-2x}(A\cos 3x+B\sin 3x)\).
Conditions: \(y(0)=2\) gives \(A=2\). Differentiating by the product rule, \(y'=e^{-2x}\big((3B-2A)\cos 3x-(3A+2B)\sin 3x\big)\), so \(y'(0)=3B-2A=1\) gives \(B=\frac{5}{3}\).
Particular solution: \(y=e^{-2x}\left(2\cos 3x+\tfrac{5}{3}\sin 3x\right)\).
MechanismNon-homogeneous: complementary function plus particular integral
When the right side is non-zero — \(y''+ay'+by=f(x)\) — the full solution is a sum of two parts: the complementary function (the general solution of the homogeneous version, carrying both arbitrary constants) plus one particular integral, any single function that actually produces \(f(x)\). The logic is worth one sentence of understanding: the CF is everything the system does on its own; the PI is one response to the forcing; their sum is the general solution because the difference of any two solutions solves the homogeneous equation.
The PI is found by trialling a function shaped like \(f(x)\) with unknown coefficients: for a polynomial, a general polynomial of the same degree; for \(ke^{px}\), try \(\lambda e^{px}\); for \(\cos\omega x\) or \(\sin\omega x\), try \(\lambda\cos\omega x+\mu\sin\omega x\) — both terms, even if only one appears, because differentiation mixes them. The one trap: if your trial form already lives inside the CF, substituting it gives \(0\) and the method stalls. The fix is to multiply the trial by \(x\) (and by \(x^2\) against a repeated root). This is not an algebraic dodge — it is resonance, the Millennium Bridge case: forcing a system at its own natural frequency produces a response whose amplitude grows with \(x\), and the mathematics tells you so before the deck does.
Apply boundary conditions last, to the complete solution CF + PI — never to the CF alone, because the PI shifts the curve the constants must then fit.
Solve \(y''-3y'+2y=e^{3x}\).
Complementary function: auxiliary equation \(m^2-3m+2=0\), so \((m-1)(m-2)=0\), \(m=1,2\), giving \(y_{CF}=Ae^{x}+Be^{2x}\).
Particular integral: \(e^{3x}\) is not in the CF, so try \(y=\lambda e^{3x}\). Then \(y'=3\lambda e^{3x}\), \(y''=9\lambda e^{3x}\), and substituting: \((9-9+2)\lambda e^{3x}=e^{3x}\), so \(2\lambda=1\), \(\lambda=\frac{1}{2}\).
General solution: \(y=Ae^{x}+Be^{2x}+\frac{1}{2}e^{3x}\).
Had the right side been \(e^{2x}\) instead — a term already in the CF — the trial \(\lambda e^{2x}\) would collapse to \(0=e^{2x}\); the correct trial is \(\lambda xe^{2x}\), which yields \(\lambda=1\).
CaseSHM and damping: building the bridge equation
This is where modelling (spec I3) meets the solving machinery. Hang a particle of mass \(m\) on a spring obeying Hooke's law \(T=kx\), displace it, and Newton's second law along the motion reads \(m\ddot{x}=-kx\): the restoring force opposes displacement. Divide by \(m\) and you have simple harmonic motion, \(\ddot{x}=-\omega^2x\) with \(\omega^2=\frac{k}{m}\). Its auxiliary equation \(m^2+\omega^2=0\) has purely imaginary roots \(\pm\omega i\), so \(x=A\cos\omega t+B\sin\omega t\) — equivalently \(R\sin(\omega t+\phi)\) — oscillation that never dies, with period \(\frac{2\pi}{\omega}\) and amplitude \(R=\sqrt{A^2+B^2}\). Interpretation is examinable: maximum speed \(R\omega\) at the centre, maximum acceleration \(R\omega^2\) at the extremes.
Real systems lose energy. Add a resistive force proportional to velocity — magnitude \(c\dot{x}\), always opposing motion — and the model becomes \(m\ddot{x}+c\dot{x}+kx=0\). Now the discriminant \(c^2-4mk\) classifies the physics. Light damping (\(c^2<4mk\)): complex roots, \(x=e^{-\frac{c}{2m}t}(A\cos\beta t+B\sin\beta t)\) — oscillations inside a decaying exponential envelope; the pre-fix Millennium Bridge, a struck tuning fork. Critical damping (\(c^2=4mk\)): repeated root, \((A+Bt)e^{mt}\) — the fastest return to equilibrium with no oscillation, which is what car suspension and door closers are tuned to. Heavy damping (\(c^2>4mk\)): two negative real roots, a slow ooze back with no oscillation at all. Arup's dampers pushed the bridge's effective \(c\) up so that pedestrian forcing could no longer sustain a growing oscillation — a £5 million change to one coefficient's side of an inequality.
A particle moves so that \(\ddot{x}+2\dot{x}+5x=0\), with \(x=1\) and \(\dot{x}=0\) at \(t=0\). Classify the damping and find \(x(t)\).
Classify: the discriminant is \(2^2-4(5)=-16<0\): light damping — the particle oscillates with decaying amplitude.
Solve: \(m^2+2m+5=0\) gives \(m=-1\pm 2i\), so \(x=e^{-t}(A\cos 2t+B\sin 2t)\). From \(x(0)=1\), \(A=1\). Then \(\dot{x}=e^{-t}\big((2B-A)\cos 2t-(2A+B)\sin 2t\big)\), and \(\dot{x}(0)=2B-1=0\) gives \(B=\frac{1}{2}\).
Answer: \(x=e^{-t}\left(\cos 2t+\tfrac{1}{2}\sin 2t\right)\) — oscillation at angular frequency \(2\), amplitude crushed by the factor \(e^{-t}\).
MechanismCoupled systems: two populations, one equation
The Hudson's Bay Company kept pelt records of Canadian lynx and snowshoe hare for the best part of a century, and the two populations rise and fall in linked cycles roughly a decade long: hares boom, lynx feast and multiply, hares crash, lynx starve, repeat. The natural model is a coupled system — two dependent variables, each one's rate of change depending on both: \(\frac{dx}{dt}=ax+by\), \(\frac{dy}{dt}=cx+dy\). Neither equation can be solved alone, because each contains the other's variable.
AQA's method is elimination, and it is entirely mechanical. Rearrange the first equation to express \(y\) in terms of \(x\) and \(\dot{x}\); differentiate it to get \(\dot{y}\) in terms of \(\dot{x}\) and \(\ddot{x}\); substitute both into the second equation. Everything in \(y\) vanishes and a single second-order equation in \(x\) remains — which the auxiliary-equation machinery above solves on sight. Recover \(y\) at the end from your rearrangement (never by integrating from scratch, which introduces a spurious extra constant). The character of the roots again narrates the ecology: complex roots mean the populations cycle, as the pelt data shows; real negative roots mean both settle to equilibrium.
Solve \(\dot{x}=4x-2y\), \(\dot{y}=3x-y\).
Eliminate: from the first equation, \(y=\frac{4x-\dot{x}}{2}\), so \(\dot{y}=\frac{4\dot{x}-\ddot{x}}{2}\). Substitute both into the second: \(\frac{4\dot{x}-\ddot{x}}{2}=3x-\frac{4x-\dot{x}}{2}\). Multiplying by \(2\): \(4\dot{x}-\ddot{x}=6x-4x+\dot{x}\), which tidies to \(\ddot{x}-3\dot{x}+2x=0\).
Solve: auxiliary equation \(m^2-3m+2=0\), roots \(m=1,2\), so \(x=Ae^{t}+Be^{2t}\).
Recover \(y\): \(\dot{x}=Ae^{t}+2Be^{2t}\), so \(y=\frac{4x-\dot{x}}{2}=\frac{3Ae^{t}+2Be^{2t}}{2}\). Two constants in total — exactly as a two-variable first-order system demands.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
The examiner tracks the skeleton of your method, so make each stage visible: for first-order, state \(P(x)\), show the integrating factor calculation, and write the collapsed form \(\frac{d}{dx}(\mu y)=\mu Q\) explicitly before integrating — jumping straight to the answer forfeits method marks if a slip appears later. For second-order, write the auxiliary equation as its own line, state its roots, and only then write the solution form; when the question says 'hence' after a discriminant part, it is telling you the classification is worth a mark on its own.
Modelling questions reward the setup as much as the solving. When forming an equation from Hooke's law or a resistance proportional to velocity, define positive direction first, then write Newton's second law with every force signed against it — mark schemes routinely give two marks for a correct \(m\ddot{x}=-kx-c\dot{x}\) with directions justified. Classify damping by computing \(c^2-4mk\) and comparing, with the numbers shown. And in every applied answer, close with one sentence of interpretation in context — 'the oscillations decay, so the population cycles die out' — because the interpretation mark exists on almost every modelling question and is the single most-dropped mark in this topic. Where a condition like \(\dot{x}(0)=0\) is given, differentiate your general solution carefully with the product rule before substituting; the commonest algebra error in the topic is differentiating \(e^{\alpha t}(A\cos\beta t+B\sin\beta t)\) and losing a term.