AS & A Level · Practice questions · Second order: complementary functions

Three cases, one quadratic.

Six original Cambridge-style questions on complementary functions: the three cases of the auxiliary equation, finding the arbitrary constants from initial conditions, and describing the motion that a solution represents.

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

Auxiliary equation, roots, then the matching form.

01
Recall
[2 marks]

Find the general solution of \(\dfrac{d^{2}y}{dx^{2}} - 7\dfrac{dy}{dx} + 10y = 0\).

  • \(m^{2} - 7m + 10 = (m - 2)(m - 5) = 0\), so \(m = 2\) or \(5\). ✓
  • \(y = Ae^{2x} + Be^{5x}\). ✓
02
Calculation
[3 marks]

Find the general solution of \(9\dfrac{d^{2}y}{dx^{2}} - 12\dfrac{dy}{dx} + 4y = 0\).

  • \(9m^{2} - 12m + 4 = 0\); \(b^{2} - 4ac = 144 - 144 = 0\), so the root is repeated. ✓
  • \((3m - 2)^{2} = 0\), so \(m = \tfrac{2}{3}\). ✓
  • \(y = (A + Bx)e^{\frac{2}{3}x}\). ✓
03
Calculation
[3 marks]

Find the general solution of \(\dfrac{d^{2}y}{dx^{2}} + 6\dfrac{dy}{dx} + 34y = 0\).

  • \(m^{2} + 6m + 34 = 0\): \[m = \frac{-6 \pm \sqrt{36 - 136}}{2} = \frac{-6 \pm 10i}{2}\] ✓
  • \(m = -3 \pm 5i\), so \(p = -3\), \(q = 5\). ✓
  • \(y = e^{-3x}(A\cos 5x + B\sin 5x)\). ✓
04
Analysis
[4 marks]

Find the solution of \(\dfrac{d^{2}y}{dx^{2}} + 2\dfrac{dy}{dx} - 3y = 0\) for which \(y = 4\) and \(\dfrac{dy}{dx} = 0\) when \(x = 0\). Describe the behaviour of \(y\) as \(x \to \infty\).

  • \(m^{2} + 2m - 3 = (m - 1)(m + 3) = 0\), so \(y = Ae^{x} + Be^{-3x}\). ✓
  • \(x = 0\): \(A + B = 4\). \(\dfrac{dy}{dx} = Ae^{x} - 3Be^{-3x}\), so \(A - 3B = 0\). ✓
  • Subtracting: \(4B = 4\), so \(B = 1\) and \(A = 3\): \(y = 3e^{x} + e^{-3x}\). ✓
  • As \(x \to \infty\), \(e^{-3x} \to 0\) and \(3e^{x} \to \infty\), so \(y\) grows without limit, behaving like \(3e^{x}\). ✓
05
Analysis
[5 marks]

The displacement \(x\) cm of a particle at time \(t\) s satisfies \(\dfrac{d^{2}x}{dt^{2}} + 4\dfrac{dx}{dt} + 20x = 0\). When \(t = 0\), \(x = 0\) and \(\dfrac{dx}{dt} = 8\).

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

(b) Describe the motion and find the time at which the particle first returns to \(x = 0\).

  • (a) \(m^{2} + 4m + 20 = 0\): \(m = \dfrac{-4 \pm \sqrt{16 - 80}}{2} = -2 \pm 4i\), so \(x = e^{-2t}(A\cos 4t + B\sin 4t)\). ✓
  • \(t = 0\), \(x = 0\): \(A = 0\), so \(x = Be^{-2t}\sin 4t\). ✓
  • \(\dfrac{dx}{dt} = Be^{-2t}(4\cos 4t - 2\sin 4t)\); at \(t = 0\): \(4B = 8\), so \(B = 2\) and \(x = 2e^{-2t}\sin 4t\). ✓
  • (b) The real part \(-2\) is negative, so the particle oscillates about \(x = 0\) with amplitude decreasing like \(2e^{-2t}\): a damped oscillation that dies away. ✓
  • \(x = 0\) again when \(\sin 4t = 0\) with \(t > 0\): \(4t = \pi\), so \(t = \dfrac{\pi}{4} \approx 0.785\) s. ✓
06
Analysis
[5 marks]

The constant \(k\) satisfies \(k \ge 0\) and \(\dfrac{d^{2}y}{dx^{2}} + k\dfrac{dy}{dx} + 9y = 0\).

(a) Find the value of \(k\) for which the auxiliary equation has a repeated root, and write down the general solution in that case.

(b) Write down the general solution when \(k = 0\) and when \(k = 10\).

(c) Show that for \(0 < k < 6\) every non-zero solution is a damped oscillation.

  • (a) \(m^{2} + km + 9 = 0\) has a repeated root when \(k^{2} - 36 = 0\); with \(k \ge 0\), \(k = 6\). ✓
  • Then \((m + 3)^{2} = 0\) and \(y = (A + Bx)e^{-3x}\). ✓
  • (b) \(k = 0\): \(m = \pm 3i\), so \(y = A\cos 3x + B\sin 3x\). \(k = 10\): \((m + 1)(m + 9) = 0\), so \(y = Ae^{-x} + Be^{-9x}\). ✓
  • (c) For \(0 < k < 6\), \(k^{2} - 36 < 0\), so \[m = -\frac{k}{2} \pm \frac{\sqrt{36 - k^{2}}}{2}\,i\] ✓
  • The roots are complex with real part \(-\tfrac{k}{2} < 0\), so \(y = e^{-\frac{k}{2}x}(A\cos qx + B\sin qx)\) with \(q = \tfrac{1}{2}\sqrt{36 - k^{2}} > 0\): an oscillation whose amplitude decays, that is, damped. ✓

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.