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AS & A Level · Practice questions · Particular integrals and particular solutions

CF first, then the PI.

Six original Cambridge-style questions on particular integrals: choosing the trial form, the clash rule, finding a coefficient in a given form, particular solutions from two initial conditions and the long-term behaviour of a modelled oscillation.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Five steps to a particular solution.

01
Recall
[3 marks]

State a suitable form of particular integral for each equation. Do not evaluate the constants.

(a) \(y'' + y' - 2y = 3x^{2} - 1\)   (b) \(y'' + 9y = 4\cos 3x\)   (c) \(y'' - 6y' + 9y = e^{3x}\)

  • (a) The CF is \(Ae^{x} + Be^{-2x}\), with no polynomial terms: try \(ax^{2} + bx + c\). ✓
  • (b) The CF is \(A\cos 3x + B\sin 3x\), which clashes with the usual trial: try \(x(C\cos 3x + D\sin 3x)\). ✓
  • (c) \((m - 3)^{2} = 0\), a repeated root, so the CF \((A + Bx)e^{3x}\) contains \(e^{3x}\) and \(xe^{3x}\): try \(Cx^{2}e^{3x}\). ✓
02
Calculation
[4 marks]

Find the general solution of \(\dfrac{d^{2}y}{dx^{2}} + 4\dfrac{dy}{dx} + 4y = 8x + 4\).

  • CF: \(m^{2} + 4m + 4 = (m + 2)^{2} = 0\), so CF \(= (A + Bx)e^{-2x}\). ✓
  • PI: try \(y = ax + b\), so \(y' = a\), \(y'' = 0\): \(4a + 4(ax + b) = 8x + 4\). ✓
  • \(x\): \(4a = 8\), \(a = 2\). Constant: \(4a + 4b = 4\), so \(8 + 4b = 4\) and \(b = -1\). ✓
  • General solution: \(y = (A + Bx)e^{-2x} + 2x - 1\). ✓
03
Calculation
[5 marks]

Find the general solution of \(\dfrac{d^{2}y}{dx^{2}} - 2\dfrac{dy}{dx} + 5y = 10\sin x\).

  • CF: \(m^{2} - 2m + 5 = 0\), \(m = 1 \pm 2i\), so CF \(= e^{x}(A\cos 2x + B\sin 2x)\). ✓
  • PI: try \(y = C\cos x + D\sin x\), so \(y' = -C\sin x + D\cos x\), \(y'' = -C\cos x - D\sin x\). ✓
  • Cosines: \(-C - 2D + 5C = 4C - 2D = 0\), so \(D = 2C\). ✓
  • Sines: \(-D + 2C + 5D = 2C + 4D = 10\), so \(2C + 8C = 10\), \(C = 1\), \(D = 2\). ✓
  • General solution: \(y = e^{x}(A\cos 2x + B\sin 2x) + \cos x + 2\sin x\). ✓
04
Analysis
[7 marks]

(a) Explain why \(Ce^{2x}\) cannot be a particular integral of \(\dfrac{d^{2}y}{dx^{2}} - \dfrac{dy}{dx} - 2y = 6e^{2x}\).

(b) Given that \(y = kxe^{2x}\) is a particular integral, find \(k\).

(c) Find the solution for which \(y = 0\) and \(\dfrac{dy}{dx} = 1\) when \(x = 0\).

  • (a) \(m^{2} - m - 2 = (m - 2)(m + 1) = 0\), so the CF is \(Ae^{2x} + Be^{-x}\). \(Ce^{2x}\) is part of the CF, so it makes the left side zero, never \(6e^{2x}\). ✓
  • (b) \(y' = k(1 + 2x)e^{2x}\), \(y'' = k(4 + 4x)e^{2x}\). ✓
  • \(y'' - y' - 2y = k\big[(4 + 4x) - (1 + 2x) - 2x\big]e^{2x} = 3ke^{2x} = 6e^{2x}\), so \(k = 2\). ✓
  • (c) \(y = Ae^{2x} + Be^{-x} + 2xe^{2x}\), so \(y' = 2Ae^{2x} - Be^{-x} + (2 + 4x)e^{2x}\). ✓
  • \(x = 0\): \(A + B = 0\) and \(2A - B + 2 = 1\), so \(2A - B = -1\). ✓
  • With \(B = -A\): \(3A = -1\), \(A = -\tfrac{1}{3}\), \(B = \tfrac{1}{3}\). ✓
  • \(y = -\tfrac{1}{3}e^{2x} + \tfrac{1}{3}e^{-x} + 2xe^{2x}\). ✓
05
Calculation
[6 marks]

Find the particular solution of \(\dfrac{d^{2}y}{dx^{2}} + 3\dfrac{dy}{dx} + 2y = 4x\) for which \(y = 0\) and \(\dfrac{dy}{dx} = 0\) when \(x = 0\). State the function that \(y\) approaches for large positive \(x\).

  • CF: \(m^{2} + 3m + 2 = 0\), \(m = -1\) or \(-2\), so CF \(= Ae^{-x} + Be^{-2x}\). ✓
  • PI: try \(ax + b\): \(3a + 2ax + 2b = 4x\), so \(a = 2\) and \(6 + 2b = 0\), \(b = -3\). ✓
  • \(y = Ae^{-x} + Be^{-2x} + 2x - 3\), \(y' = -Ae^{-x} - 2Be^{-2x} + 2\). ✓
  • \(x = 0\): \(A + B - 3 = 0\) and \(-A - 2B + 2 = 0\). Adding: \(-B - 1 = 0\), so \(B = -1\) and \(A = 4\). ✓
  • \(y = 4e^{-x} - e^{-2x} + 2x - 3\). ✓
  • As \(x \to \infty\) both exponentials tend to zero, so \(y\) approaches the line \(y = 2x - 3\) (the PI). ✓
06
Analysis
[8 marks]

A light spring with damping is driven by a periodic force. The displacement \(x\) metres at time \(t\) seconds satisfies \[\dfrac{d^{2}x}{dt^{2}} + 2\dfrac{dx}{dt} + 2x = 10\cos 2t,\] and the system starts at rest with \(x = 0\).

(a) Find the complementary function.

(b) Find a particular integral.

(c) Find \(x\) in terms of \(t\).

(d) Describe the motion for large \(t\), stating its amplitude, and explain why the initial conditions do not affect it.

  • (a) \(m^{2} + 2m + 2 = 0\), \(m = -1 \pm i\), so CF \(= e^{-t}(A\cos t + B\sin t)\). ✓
  • (b) Try \(C\cos 2t + D\sin 2t\). Cosines: \(-4C + 4D + 2C = -2C + 4D = 10\). Sines: \(-4D - 4C + 2D = -4C - 2D = 0\). ✓
  • So \(D = -2C\), \(-2C - 8C = 10\): \(C = -1\), \(D = 2\). PI \(= -\cos 2t + 2\sin 2t\). ✓
  • (c) \(x = e^{-t}(A\cos t + B\sin t) - \cos 2t + 2\sin 2t\). At \(t = 0\), \(x = 0\): \(A - 1 = 0\), \(A = 1\). ✓
  • \(\dfrac{dx}{dt}\) at \(t = 0\) is \(-A + B + 4 = 0\) (the PI contributes \(2D = 4\)), so \(B = -3\). ✓
  • \[\begin{aligned} x = {}&e^{-t}(\cos t - 3\sin t) \\ &- \cos 2t + 2\sin 2t. \end{aligned}\] ✓
  • (d) \(e^{-t} \to 0\), so for large \(t\), \(x \approx -\cos 2t + 2\sin 2t\): an oscillation of period \(\pi\) s and amplitude \(\sqrt{1 + 4} = \sqrt{5} \approx 2.24\) m. ✓
  • The initial conditions only fix \(A\) and \(B\), which multiply \(e^{-t}\) and so die away (by \(t = 5\) the transient is below \(\sqrt{10}\,e^{-5} \approx 0.02\) m). The PI contains no arbitrary constants. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.