AS & A Level · Practice questions · Solving by separating the variables

Sort, integrate, substitute.

Six original Cambridge-style questions on solving differential equations by separating the variables, with initial conditions and interpretation in context.

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

Separate, integrate, constant, rearrange.

Separate: \(\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\); split \(e^{a + b}\) into \(e^{a}e^{b}\) first.
Integrate both sides, using partial fractions or standard integrals as needed.
One constant, found from the initial condition straight away.
Rearrange to the form asked for; exponentiating turns \(+C\) into a multiplier.
Interpret: answer in context with units, and say what happens as \(t \to \infty\).

Recall

[3 marks]

Find the general solution of \(\dfrac{dy}{dx} = \dfrac{2y}{x}\) for \(x > 0\), \(y > 0\), giving \(y\) in terms of \(x\).

\(\displaystyle\int \frac{1}{y}\,dy = \int \frac{2}{x}\,dx\), so \(\ln y = 2\ln x + C\). ✓
\(y = e^{C}e^{2\ln x} = Ax^{2}\), where \(A = e^{C}\). ✓

Calculation

[4 marks]

Solve \(\dfrac{dx}{dt} = \cos^{2}x\), given that \(x = 0\) when \(t = 0\). Find the value of \(t\) when \(x = \tfrac{\pi}{4}\).

\(\displaystyle\int \sec^{2}x\,dx = \int dt\), so \(\tan x = t + C\). ✓
\(x = 0\), \(t = 0\): \(C = 0\), so \(\tan x = t\), i.e. \(x = \tan^{-1} t\). ✓
\(x = \tfrac{\pi}{4}\): \(t = \tan\tfrac{\pi}{4} = 1\). ✓

Calculation

[4 marks]

Solve \(\dfrac{dy}{dx} = x(y^{2} + 1)\), given that \(y = 1\) when \(x = 0\), giving \(y\) in terms of \(x\).

\(\displaystyle\int \frac{1}{y^{2} + 1}\,dy = \int x\,dx\), so \(\tan^{-1} y = \tfrac{1}{2}x^{2} + C\). ✓
\(x = 0\), \(y = 1\): \(C = \tfrac{\pi}{4}\). ✓
\(y = \tan\left(\tfrac{1}{2}x^{2} + \tfrac{\pi}{4}\right)\). ✓

Analysis

[5 marks]

A metal bar cools according to \(\dfrac{d\theta}{dt} = -0.05(\theta - 18)\), where \(\theta\) is in °C and \(t\) in minutes. Initially \(\theta = 98\). (a) Find \(\theta\) in terms of \(t\). (b) Find the time taken for the bar to cool to \(38^\circ\)C. (c) State the temperature the bar approaches.

(a) \(\displaystyle\int \frac{1}{\theta - 18}\,d\theta = \int -0.05\,dt\): \(\ln(\theta - 18) = -0.05t + C\), \(C = \ln 80\). ✓
\(\theta = 18 + 80e^{-0.05t}\). ✓
(b) \(80e^{-0.05t} = 20\), so \(e^{-0.05t} = \tfrac{1}{4}\) and \(t = 20\ln 4 = 27.7\) minutes (3 s.f.). ✓
(c) \(e^{-0.05t} \to 0\), so \(\theta \to 18^\circ\)C, the room temperature. ✓

Analysis

[7 marks]

The number of people \(x\) (in thousands) who have heard a rumour satisfies \(\dfrac{dx}{dt} = \dfrac{x(6 - x)}{12}\), where \(t\) is in days. Initially \(x = 2\).

(a) Show that \(x = \dfrac{6}{1 + 2e^{-t/2}}\). (b) Find the time when 5000 people have heard the rumour. (c) State what happens to \(x\) in the long term.

(a) \(\displaystyle\int \frac{12}{x(6 - x)}\,dx = \int dt\), with \(\dfrac{12}{x(6 - x)} = \dfrac{2}{x} + \dfrac{2}{6 - x}\). ✓
\(2\ln x - 2\ln(6 - x) = t + C\); \(t = 0\), \(x = 2\): \(C = 2\ln\tfrac{2}{4} = -2\ln 2\). ✓
\(2\ln\dfrac{2x}{6 - x} = t\), so \(\dfrac{2x}{6 - x} = e^{t/2}\), \(2x = 6e^{t/2} - xe^{t/2}\), \(x = \dfrac{6e^{t/2}}{2 + e^{t/2}} = \dfrac{6}{1 + 2e^{-t/2}}\). ✓
(b) \(x = 5\): \(\dfrac{10}{1} = e^{t/2}\), so \(t = 2\ln 10 = 4.61\) days (3 s.f.). ✓
(c) \(e^{-t/2} \to 0\), so \(x \to 6\): about 6000 people hear the rumour in the long run. ✓

Analysis

[6 marks]

The depth \(h\) metres of water in a draining tank satisfies \(\dfrac{dh}{dt} = -0.02\sqrt{h}\), where \(t\) is in minutes. Initially \(h = 0.81\).

(a) Solve the differential equation, giving \(h\) in terms of \(t\). (b) Find the depth after 30 minutes. (c) Find the time taken for the tank to empty.

(a) \(\displaystyle\int h^{-1/2}\,dh = \int -0.02\,dt\): \(2\sqrt{h} = -0.02t + C\). ✓
\(t = 0\): \(C = 2\sqrt{0.81} = 1.8\), so \(\sqrt{h} = 0.9 - 0.01t\) and \(h = (0.9 - 0.01t)^{2}\). ✓
(b) \(t = 30\): \(h = 0.6^{2} = 0.36\) m. ✓
(c) Empty when \(\sqrt{h} = 0\): \(0.9 - 0.01t = 0\), so \(t = 90\) 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.