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  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn7367Paper 2 · Compulsory contentDifferential equationsSecond-order equations: auxiliary equation
Differential equations · Subtopic 2 of 2

Second-Order Linear Differential Equations

For ay′′+by′+cy=f(x)ay''+by'+cy=f(x), try y=emxy=e^{mx} in the "homogeneous" equation: am2+bm+c=0am^2+bm+c=0 (the auxiliary equation). Its roots give the complementary function. Add a particular integral matching f(x)f(x), then fix the two constants with two conditions.

Cambridge 9231 Paper 2Edexcel YFM01 WFM02Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 50 minRevision 1 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Form and solve the auxiliary equation.
  • Write complementary functions for distinct real, repeated and complex roots.
  • Find particular integrals for polynomial, exponential and trigonometric right-hand sides.
  • Use two conditions to find a particular solution.
Pause and think

Which function is its own derivative up to a constant multiple?

Section 2 of 14

Core idea

Start hereBouncing on a spring

A mass on a spring with a damper obeys mx¨+cx˙+kx=0m\ddot x+c\dot x+kx=0: acceleration depends on position and velocity. Will it oscillate, or creep back to rest? The answer depends on the roots of an ordinary quadratic.

The big idea

ay′′+by′+cy=f(x)ay''+by'+cy=f(x). Auxiliary equation am2+bm+c=0am^2+bm+c=0 gives the complementary function; add a particular integral for f(x)f(x); use two conditions for the constants.

1. The auxiliary equation

Trying y=emxy=e^{mx}
  1. y=emx, y′=memx, y′′=m2emxy=e^{mx},\ y'=me^{mx},\ y''=m^2e^{mx}
    Exponentials keep their shape.
  2. (am2+bm+c)emx=0(am^2+bm+c)e^{mx}=0
    Substitute into ay′′+by′+cy=0ay''+by'+cy=0.
  3. am2+bm+c=0am^2+bm+c=0
    emx≠0e^{mx}\neq0.
Each root gives a solution; linear combinations are also solutions.
Complex roots become trig
  1. m=p±qim=p\pm qi
    Negative discriminant.
  2. y=epx(Ceiqx+De−iqx)y=e^{px}(Ce^{iqx}+De^{-iqx})
    Two complex exponential solutions.
  3. =epx(Acos⁡qx+Bsin⁡qx)=e^{px}(A\cos qx+B\sin qx)
    Euler: eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta; regroup constants.
CF={Aeαx+Beβx(A+Bx)eαxepx(Acos⁡qx+Bsin⁡qx)\text{CF}=\begin{cases}Ae^{\alpha x}+Be^{\beta x}\\(A+Bx)e^{\alpha x}\\e^{px}(A\cos qx+B\sin qx)\end{cases}
where
α,β\alpha,\beta
distinct real roots.
α\alpha (repeated)
the extra xx gives a second independent solution.
p±qip\pm qi
complex roots.

2. Particular integrals

Worked exampley′′−3y′+2y=4y''-3y'+2y=4

General solution.

Goal
yy.
Strategy
CF from m2−3m+2=0m^2-3m+2=0; PI a constant.
  1. m=1,2⇒CF=Aex+Be2xm=1,2\Rightarrow\text{CF}=Ae^x+Be^{2x}
    Factorise (m−1)(m−2)(m-1)(m-2).
  2. y=λ⇒2λ=4⇒λ=2y=\lambda\Rightarrow2\lambda=4\Rightarrow\lambda=2
    Derivatives of a constant vanish.
Answery=Aex+Be2x+2y=Ae^x+Be^{2x}+2.
✓ Checky=2y=2: 0−0+4=40-0+4=4 ✓.
Worked exampley′′−y=0y''-y=0, y(0)=2y(0)=2, y′(0)=0y'(0)=0

Particular solution.

Goal
yy.
Strategy
CF; two equations for AA, BB.
  1. y=Aex+Be−xy=Ae^x+Be^{-x}
    m=±1m=\pm1.
  2. A+B=2, A−B=0A+B=2,\ A-B=0
    y(0)y(0) and y′(0)y'(0).
  3. A=B=1A=B=1
    Solve.
Answery=ex+e−x=2cosh⁡xy=e^x+e^{-x}=2\cosh x.
✓ Checky′′=yy''=y ✓.
Common misconceptionPI already in the CF If f(x)=exf(x)=e^{x} and 11 is a root, try λxex\lambda xe^x (or λx2ex\lambda x^2e^x for a repeated root).
Key takeawayDamping For x¨+2kx˙+ω2x=0\ddot x+2k\dot x+\omega^2x=0: k<ωk<\omega light (oscillates), k=ωk=\omega critical, k>ωk>\omega heavy.
Pause and think

What does y′′+9y=0y''+9y=0 describe?

Summary

  • Auxiliary equation from emxe^{mx}.
  • Three CF cases.
  • PI from the form of f(x)f(x).
  • GS = CF + PI; two conditions.
Key words
Complementary function
General solution with f(x)=0f(x)=0.
Particular integral
One solution with the given f(x)f(x).
HistoryEuler. Leonhard Euler solved linear equations with constant coefficients in 1743, introducing the trial solution emxe^{mx}.
Fun factCar suspension. A car's springs and dampers obey my′′+cy′+ky=0my''+cy'+ky=0. Engineers tune cc so the auxiliary equation has a repeated root: critical damping, the smoothest ride.
Section 3 of 14

Key formulae

Auxiliary equationam2+bm+c=0am^2+bm+c=0
Complementary function{Aeαx+Beβxdistinct real(A+Bx)eαxrepeatedepx(Acos⁡qx+Bsin⁡qx)m=p±qi\begin{cases}Ae^{\alpha x}+Be^{\beta x}&\text{distinct real}\\(A+Bx)e^{\alpha x}&\text{repeated}\\e^{px}(A\cos qx+B\sin qx)&m=p\pm qi\end{cases}
General solutiony=CF+PIy=\text{CF}+\text{PI}
TrickTrial PIs. f(x)=kf(x)=k: try λ\lambda. Linear: λx+μ\lambda x+\mu. kenxke^{nx}: λenx\lambda e^{nx} (or λxenx\lambda xe^{nx} if nn is a root). sin⁡\sin/cos⁡\cos: λcos⁡+μsin⁡\lambda\cos+\mu\sin.
Section 4 of 14

Worked examples

Example 1 · Distinct roots0 / 3 steps
Solve y′′−3y′+2y=4y''-3y'+2y=4.
    Example 2 · Complex roots0 / 2 steps
    Solve y′′+2y′+5y=0y''+2y'+5y=0.
      Section 5 of 14

      Explore the graph

      Damped oscillation
      Loading the interactive graph…
      Try this
      1. Roots −p±qi-p\pm qi give e−pxcos⁡qxe^{-px}\cos qx: oscillation inside a decaying envelope.
      2. Set p=0p=0: undamped SHM.

      Complex roots: the real part sets decay, the imaginary part sets frequency.

      In the real worldElectrical circuits. An inductor, resistor and capacitor in series satisfy Lq′′+Rq′+qC=V(t)Lq''+Rq'+\frac qC=V(t): the same equation as a damped spring.
      Section 6 of 14

      Common mistakes

      IncorrectPI λex\lambda e^{x} when 11 is a root
      CorrectTry λxex\lambda xe^x
      λex\lambda e^x is already in the CF, so it gives zero.
      IncorrectApplying conditions to the CF only
      CorrectUse the full general solution
      The PI contributes to yy and y′y'.
      Incorrecte±qie^{\pm qi} left in the answer
      CorrectWrite Acos⁡qx+Bsin⁡qxA\cos qx+B\sin qx
      Real solutions are expected.

      Second-order ODE: looks terrifying

      Auxiliary equation: it's just a quadratic, mate.

      Section 7 of 14

      Exam tip

      Exam tipTwo constants, two conditions. Differentiate the general solution before substituting y′(0)y'(0); find the PI first so it's included.
      Section 8 of 14

      Guided practice

      SA-FP-DE2-G012 marksL1 FoundationNon-calculatorComplementary function
      Find the roots of the auxiliary equation for y′′−5y′+6y=0y''-5y'+6y=0.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-DE2-0012 marksL2 StandardNon-calculatorComplementary function
      The general solution of y′′+4y=0y''+4y=0 is Acos⁡ωx+Bsin⁡ωxA\cos\omega x+B\sin\omega x. Find ω\omega.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-DE2-0021 markL2 StandardNon-calculatorComplementary function
      Find the repeated root of the auxiliary equation for y′′+2y′+y=0y''+2y'+y=0.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-DE2-0032 marksL2 StandardNon-calculatorParticular integral
      Find the constant particular integral of y′′−3y′+2y=4y''-3y'+2y=4.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-DE2-0043 marksL3 ExamNon-calculatorParticular integral
      Find a particular integral of y′′+y=3xy''+y=3x, in terms of xx.

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-DE2-0055 marksL4 ChallengingNon-calculatorParticular solutions
      y′′−y=0y''-y=0, y(0)=2y(0)=2, y′(0)=0y'(0)=0. Find the exact value of yy when x=ln⁡2x=\ln2.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • Auxiliary equation am2+bm+c=0am^2+bm+c=0.
      • Three CF forms.
      • PI matches f(x)f(x).
      • GS = CF + PI; two conditions.
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-DE2-Q011 markL1 FoundationNon-calculatorComplementary function
      If the auxiliary equation has roots p±qip\pm qi, the complementary function is
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      7367 A-level Further Mathematics
      Where
      Paper 1 and 2 · Differential equations
      Expected
      Second-order equations with constant coefficients; SHM and damping.
      Calculator
      Permitted on all papers.
      Typical style
      AQA mixes short multiple-choice questions with longer structured ones.
      Open in
      7367 · Paper 17367 · Paper 2
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

      Browse past papers

      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).