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How do you solve first and second order differential equations, and how do they model simple harmonic motion and damping?

First order linear differential equations by the integrating factor, second order linear constant-coefficient equations via the auxiliary equation and particular integral, and applications to simple harmonic motion and damped systems.

A focused answer to the OCR A-Level Further Mathematics A content on differential equations, covering first order linear equations by the integrating factor, second order constant-coefficient equations via the auxiliary equation and a particular integral, and applications to simple harmonic motion and damped systems.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. First order linear equations: the integrating factor
  3. Second order equations: complementary function
  4. Particular integral and the general solution
  5. Simple harmonic motion and damping
  6. Try this

What this dot point is asking

OCR wants you to solve first order linear differential equations using an integrating factor, solve second order linear constant-coefficient equations by forming the auxiliary equation for the complementary function and adding a particular integral for a non-zero right-hand side, and interpret second order equations as models of simple harmonic motion and of damped systems, classifying the damping from the nature of the roots.

First order linear equations: the integrating factor

A first order equation is linear when it can be written dydx+P(x)y=Q(x)\dfrac{dy}{dx} + P(x)y = Q(x). Multiplying by the integrating factor turns the left-hand side into the derivative of a product, which you can then integrate directly.

Second order equations: complementary function

A homogeneous second order equation ax¨+bx˙+cx=0a\ddot{x} + b\dot{x} + cx = 0 is solved through its auxiliary equation am2+bm+c=0am^2 + bm + c = 0. The form of the solution depends on the roots.

Particular integral and the general solution

When the right-hand side is non-zero, ax¨+bx˙+cx=f(t)a\ddot{x} + b\dot{x} + cx = f(t), add a particular integral (a trial function shaped like ff, with coefficients fixed by substitution) to the complementary function. The full general solution is the sum, and any initial conditions then determine AA and BB.

Simple harmonic motion and damping

The equation x¨+ω2x=0\ddot{x} + \omega^2 x = 0 has auxiliary roots ±ωi\pm \omega i, giving x=Acosωt+Bsinωtx = A\cos\omega t + B\sin\omega t: simple harmonic motion of angular frequency ω\omega. Adding a resistance term bx˙b\dot{x} models damping, and the auxiliary roots classify it.

  • Light (under) damping. Complex roots p±qip \pm qi with p<0p < 0: the system oscillates with exponentially decaying amplitude.
  • Critical damping. A repeated real root: the fastest return to rest without oscillating.
  • Heavy (over) damping. Two distinct negative real roots: a slow return to rest with no oscillation.

Differential equations draw on integration, on complex numbers (for the oscillatory case), and connect Pure Core to the mechanics of oscillating and damped systems.

Try this

Q1. State the integrating factor for dydx+3y=x\dfrac{dy}{dx} + 3y = x. [1 mark]

  • Cue. I=e3dx=e3xI = e^{\int 3\,dx} = e^{3x}.

Q2. Write the general solution of x¨+9x=0\ddot{x} + 9x = 0. [2 marks]

  • Cue. Auxiliary roots ±3i\pm 3i, so x=Acos3t+Bsin3tx = A\cos 3t + B\sin 3t (SHM with ω=3\omega = 3).

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20196 marksSolve the differential equation dydx+2y=ex\dfrac{dy}{dx} + 2y = e^{x}, given that y=1y = 1 when x=0x = 0.
Show worked answer →

This is first order linear, so use an integrating factor (M1): I=e2dx=e2xI = e^{\int 2\,dx} = e^{2x} (A1).

Multiply through (M1): e2xdydx+2e2xy=e3xe^{2x}\dfrac{dy}{dx} + 2e^{2x}y = e^{3x}, so ddx(e2xy)=e3x\dfrac{d}{dx}\left(e^{2x}y\right) = e^{3x}.

Integrate (A1): e2xy=13e3x+Ce^{2x}y = \dfrac{1}{3}e^{3x} + C, so y=13ex+Ce2xy = \dfrac{1}{3}e^{x} + Ce^{-2x} (A1).

Apply y(0)=1y(0) = 1: 1=13+C1 = \tfrac{1}{3} + C, so C=23C = \tfrac{2}{3}, giving y=13ex+23e2xy = \tfrac{1}{3}e^{x} + \tfrac{2}{3}e^{-2x} (A1).

Markers reward the integrating factor, recognising the product derivative, integrating, and applying the condition.

OCR 20226 marksFind the general solution of d2xdt2+4dxdt+13x=0\dfrac{d^2 x}{dt^2} + 4\dfrac{dx}{dt} + 13x = 0, and state whether the motion is lightly, critically or heavily damped.
Show worked answer →

Form the auxiliary equation (M1): m2+4m+13=0m^2 + 4m + 13 = 0.

Solve (M1): m=4±16522=4±362=2±3im = \dfrac{-4 \pm \sqrt{16 - 52}}{2} = \dfrac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i (A1).

Complex roots give an oscillating, decaying solution (M1): x=e2t(Acos3t+Bsin3t)x = e^{-2t}(A\cos 3t + B\sin 3t) (A1).

Because the roots are complex (oscillation present) with a negative real part, the motion is lightly damped (under-damped): it oscillates while the amplitude decays (A1).

Markers reward the auxiliary equation, the complex roots, the correct form of the solution, and the light-damping classification.

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