Home · A-Level Further Maths 9231 · Differential equations · First order: integrating factors · Practice
AS & A Level · Practice questions · First order: integrating factors

One multiplier, one integration.

Six original Cambridge-style questions on first order linear differential equations: general solutions with an integrating factor, particular solutions, a solution that stays finite, and two modelled contexts with interpretation.

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

Standard form, factor, integrate, divide.

01
Recall
[3 marks]

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

  • \(I = e^{\int 5\,dx} = e^{5x}\), so \(\dfrac{d}{dx}\big(ye^{5x}\big) = 10e^{6x}\). ✓
  • Integrate: \(ye^{5x} = \tfrac{5}{3}e^{6x} + C\). ✓
  • Divide by \(e^{5x}\): \(y = \tfrac{5}{3}e^{x} + Ce^{-5x}\). ✓
02
Calculation
[5 marks]

Find the particular solution of \(x\dfrac{dy}{dx} + 3y = 10x^{2}\), for \(x \gt 0\), given that \(y = 5\) when \(x = 1\).

  • Standard form: \(\dfrac{dy}{dx} + \dfrac{3}{x}\,y = 10x\). ✓
  • \(I = e^{3\ln x} = x^{3}\), so \(\dfrac{d}{dx}\big(x^{3}y\big) = 10x^{4}\). ✓
  • Integrate: \(x^{3}y = 2x^{5} + C\). ✓
  • \(x = 1\), \(y = 5\): \(5 = 2 + C\), so \(C = 3\). ✓
  • \(y = 2x^{2} + \dfrac{3}{x^{3}}\). ✓
03
Calculation
[6 marks]

Solve \(\dfrac{dy}{dx} - y\tan x = 2\sin x\) for \(-\tfrac{\pi}{2} \lt x \lt \tfrac{\pi}{2}\), given that \(y = 1\) when \(x = 0\). Give \(y\) in terms of \(x\) and find the exact value of \(y\) when \(x = \tfrac{\pi}{3}\).

  • \(P = -\tan x\), so \(\int P\,dx = \ln\cos x\) and \(I = \cos x\) (positive on this interval). ✓
  • \(\dfrac{d}{dx}\big(y\cos x\big) = 2\sin x\cos x\). ✓
  • Integrate (\(\dfrac{d}{dx}\sin^{2}x = 2\sin x\cos x\)): \(y\cos x = \sin^{2}x + C\). ✓
  • \(x = 0\), \(y = 1\): \(1 = 0 + C\), so \(C = 1\) and \(y = \dfrac{1 + \sin^{2}x}{\cos x} = (1 + \sin^{2}x)\sec x\). ✓
  • At \(x = \tfrac{\pi}{3}\): \(\sin^{2}x = \tfrac{3}{4}\) and \(\sec x = 2\), so \(y = \tfrac{7}{4} \times 2 = \tfrac{7}{2}\). ✓
  • (Integrating \(\sin 2x\) as \(-\tfrac{1}{2}\cos 2x\) gives an equivalent answer with a different constant.) ✓
04
Analysis
[6 marks]

(a) Find the general solution of \(\dfrac{dy}{dx} + 2y\coth x = \cosh x\) for \(x \gt 0\).

(b) Show that exactly one solution stays finite as \(x \to 0^{+}\), and find the value of \(y\) for that solution when \(x = \ln 2\).

  • (a) \(I = e^{2\ln\sinh x} = \sinh^{2}x\) (since \(\sinh x \gt 0\) for \(x \gt 0\)). ✓
  • \(\dfrac{d}{dx}\big(y\sinh^{2}x\big) = \sinh^{2}x\cosh x\), so \(y\sinh^{2}x = \tfrac{1}{3}\sinh^{3}x + C\). ✓
  • \(y = \tfrac{1}{3}\sinh x + \dfrac{C}{\sinh^{2}x} = \tfrac{1}{3}\sinh x + C\operatorname{cosech}^{2}x\). ✓
  • (b) As \(x \to 0^{+}\), \(\tfrac{1}{3}\sinh x \to 0\) but \(\operatorname{cosech}^{2}x \to \infty\), so \(y\) stays finite only when \(C = 0\): the solution \(y = \tfrac{1}{3}\sinh x\). ✓
  • \(\sinh(\ln 2) = \tfrac{1}{2}\big(2 - \tfrac{1}{2}\big) = \tfrac{3}{4}\), so \(y = \tfrac{1}{3} \times \tfrac{3}{4} = \tfrac{1}{4}\). ✓
05
Modelling
[7 marks]

A tank of capacity 500 litres initially holds 200 litres of pure water. Brine containing 0.05 kg of salt per litre flows in at 4 litres per minute, and the well-stirred mixture flows out at 2 litres per minute. The mass of salt in the tank after \(t\) minutes is \(S\) kg.

(a) Show that \(\dfrac{dS}{dt} + \dfrac{S}{100 + t} = 0.2\).

(b) Find \(S\) in terms of \(t\).

(c) Find the concentration of salt at the moment the tank starts to overflow, and comment on the concentration if the tank were large enough never to overflow.

  • (a) Volume \(V = 200 + 2t\). Salt in: \(4 \times 0.05 = 0.2\) kg per minute. Salt out: \(2 \times \dfrac{S}{200 + 2t} = \dfrac{S}{100 + t}\). So \(\dfrac{dS}{dt} = 0.2 - \dfrac{S}{100 + t}\), which rearranges to the given equation. ✓
  • (b) \(I = e^{\ln(100 + t)} = 100 + t\), so \(\dfrac{d}{dt}\big(S(100 + t)\big) = 0.2(100 + t)\). ✓
  • Integrate: \(S(100 + t) = 0.1(100 + t)^{2} + C\). At \(t = 0\), \(S = 0\): \(0 = 1000 + C\), so \(C = -1000\). ✓
  • \(S = 0.1(100 + t) - \dfrac{1000}{100 + t}\). ✓
  • (c) The tank is full when \(200 + 2t = 500\), so \(t = 150\). Then \(S = 25 - 4 = 21\) kg and the concentration is \(\dfrac{21}{500} = 0.042\) kg per litre. ✓
  • Without overflow, \(\dfrac{S}{V} = \dfrac{S}{2(100 + t)} = 0.05 - \dfrac{500}{(100 + t)^{2}} \to 0.05\) kg per litre: the concentration would approach that of the incoming brine. ✓
06
Modelling
[8 marks]

A heater in a room is switched off at \(t = 0\), and the room temperature after \(t\) hours is then \(20 + 10e^{-t}\) °C. A metal block, initially at 50 °C, has temperature \(\theta\) °C satisfying \[\dfrac{d\theta}{dt} = -2\big(\theta - 20 - 10e^{-t}\big).\]

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

(b) Show that the block is always warmer than the room, and state its long-term temperature.

(c) Find the exact time at which \(\theta = 30\).

  • (a) Standard form: \(\dfrac{d\theta}{dt} + 2\theta = 40 + 20e^{-t}\). \(I = e^{2t}\). ✓
  • \(\dfrac{d}{dt}\big(\theta e^{2t}\big) = 40e^{2t} + 20e^{t}\), so \(\theta e^{2t} = 20e^{2t} + 20e^{t} + C\). ✓
  • \(\theta = 20 + 20e^{-t} + Ce^{-2t}\). At \(t = 0\): \(50 = 40 + C\), so \(C = 10\). \[\theta = 20 + 20e^{-t} + 10e^{-2t}.\] ✓
  • (b) \(\theta - (20 + 10e^{-t}) = 10e^{-t} + 10e^{-2t} \gt 0\) for all \(t\), so the block is always warmer than the room. As \(t \to \infty\), \(\theta \to 20\) °C, the final room temperature. ✓
  • (c) \(20e^{-t} + 10e^{-2t} = 10\). Let \(u = e^{-t}\): \(u^{2} + 2u - 1 = 0\), so \(u = -1 \pm \sqrt{2}\). ✓
  • \(u \gt 0\), so \(u = \sqrt{2} - 1\) and \(t = -\ln\big(\sqrt{2} - 1\big) = \ln\big(1 + \sqrt{2}\big) \approx 0.881\) hours (about 53 minutes). ✓

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.