Differential equations - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Differential equations for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
3–11 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Find the general solution of the differential equation
\[\dfrac{dy}{dx} = \dfrac{3}{x(y + 2)}, \quad x > 0, \; y > -2\]
giving your answer in the form \((y + 2)^2 = \mathrm{f}(x)\).

Model answer

\(\displaystyle\int (y + 2)\,dy = \int \dfrac{3}{x}\,dx\)
\(\dfrac{1}{2}(y + 2)^2 = 3\ln x + c\)
\((y + 2)^2 = 6\ln x + C\)

Explanation

Multiply by \(y + 2\) and divide by 3 over \(x\) to separate the variables; then \(\displaystyle\int (y + 2)\,dy = \dfrac{1}{2}(y + 2)^2\) and \(\displaystyle\int \dfrac{3}{x}\,dx = 3\ln x\). Doubling both sides gives the form asked, with a new constant. A common error is forgetting to double the \(3\ln x\) term, or leaving out the constant.

Mark scheme

1 mark
B1 Separates the variables correctly: \(\displaystyle\int (y + 2)\,dy = \int \dfrac{3}{x}\,dx\). A correct final answer of \((y + 2)^2 = 6\ln x + c\) with no working scores 3 marks.
An answer of \((y + 2)^2 = 6\ln x + c\) with no working scores 3 marks.
1 mark
M1 Integrates both sides: \(\alpha(y + 2)^2\) (or \(\alpha y^2 + \beta y\)) \(= \gamma\ln x\) (\(+ c\))
Not dependent on the B1.
1 mark
A1 \((y + 2)^2 = 6\ln x + c\) oe in the form asked
A constant is required. Accept \((y + 2)^2 = \ln(Ax^6)\) or \((y + 2)^2 = 6\ln x + 4 + c\) (from \(\frac{1}{2}y^2 + 2y = 3\ln x + c\)). \((y + 2)^2 = 3\ln x + c\) (not doubling the right-hand side) is A0.

Question 2

5 marks

Given that \(y = 0\) when \(x = 0\), solve the differential equation
\[\dfrac{dy}{dx} = \mathrm{e}^{y}\cos x, \quad -\dfrac{\pi}{2} < x < \dfrac{\pi}{2}\]
giving \(y\) in terms of \(x\).

Model answer

\(\dfrac{dy}{dx} = \mathrm{e}^{y}\cos x \Rightarrow \displaystyle\int \mathrm{e}^{-y}\,dy = \int \cos x\,dx\)
\(-\mathrm{e}^{-y} = \sin x + c\)
\(x = 0\), \(y = 0\): \(-1 = 0 + c\), so \(c = -1\)
\(\mathrm{e}^{-y} = 1 - \sin x\)
\(-y = \ln(1 - \sin x)\)
\(y = -\ln(1 - \sin x)\)

Explanation

Dividing by \(\mathrm{e}^{y}\) separates the variables, and \(\mathrm{e}^{-y}\) integrates to \(-\mathrm{e}^{-y}\), so watch the sign. The condition \(y = 0\) when \(x = 0\) gives \(c = -1\); then take logarithms to make \(y\) the subject. On the interval \(\sin x < 1\), so \(1 - \sin x > 0\) and the logarithm is defined. Common errors are integrating \(\mathrm{e}^{-y}\) as \(\mathrm{e}^{-y}\) and leaving the answer as \(\mathrm{e}^{-y} = 1 - \sin x\).

Mark scheme

1 mark
B1 Separates the variables correctly: \(\displaystyle\int \mathrm{e}^{-y}\,dy = \int \cos x\,dx\). A correct final answer of \(y = -\ln(1 - \sin x)\) with no working scores 2 marks (special case: M1 and dM1).
An answer of \(y = -\ln(1 - \sin x)\) (or \(y = \ln\dfrac{1}{1 - \sin x}\)) with no working scores 2 marks (special case: the M1 for integrating and the dM1 for the constant are implied). This is a multi-step problem, so a bare answer cannot score full marks.
1 mark
M1 Integrates both sides: \(\pm\mathrm{e}^{-y} = \pm\sin x\) (\(+ c\))
Integrating \(\mathrm{e}^{-y}\) to \(\mathrm{e}^{-y}\), or to a logarithm, is M0.
1 mark
A1 \(-\mathrm{e}^{-y} = \sin x + c\)
1 mark
M1 (dM1) Uses \(x = 0\), \(y = 0\) to find the constant: \(-1 = 0 + c\), \(c = -1\)
Depends on the first M mark (integrating), not on the A mark just before.
1 mark
A1 \(y = -\ln(1 - \sin x)\) oe
Accept \(y = \ln\dfrac{1}{1 - \sin x}\) or \(y = \ln\left((1 - \sin x)^{-1}\right)\). \(y\) must be the subject. \(\mathrm{e}^{-y} = 1 - \sin x\) is not enough.

Question 3

7 marks

In this question you must show all stages of your working.

A wet towel is hung on a washing line to dry. The mass of water in the towel is \(M\) grams at time \(t\) hours after it is hung out. In a model,
\[\dfrac{dM}{dt} = -\dfrac{150M}{M + 100}, \quad M > 0\]
When the towel is hung out, it contains 500 grams of water.

Show that, according to the model,
\[t = \dfrac{1}{150}\left(500 - M + 100\ln\dfrac{500}{M}\right)\]
Use the model to find how long it takes, to the nearest minute, for the mass of water in the towel to fall to 100 grams.
Explain why, according to the model, the towel never becomes completely dry.

Model answer

\(\displaystyle\int \dfrac{M + 100}{M}\,dM = \int -150\,dt\)
\(\displaystyle\int \left(1 + \dfrac{100}{M}\right) dM = -150t + c\)
\(M + 100\ln M = -150t + c\)
\(t = 0\), \(M = 500\): \(c = 500 + 100\ln 500\)
\(150t = 500 - M + 100\ln 500 - 100\ln M = 500 - M + 100\ln\dfrac{500}{M}\)
\(t = \dfrac{1}{150}\left(500 - M + 100\ln\dfrac{500}{M}\right)\)
\(M = 100\): \(t = \dfrac{1}{150}(400 + 100\ln 5) = 3.7396...\) hours \(= 3\) hours 44 minutes (to the nearest minute)
As \(M \to 0\), \(\dfrac{500}{M} \to \infty\), so \(\ln\dfrac{500}{M} \to \infty\) and \(t \to \infty\): there is no finite time at which \(M = 0\), so the towel never becomes completely dry.

Explanation

Dividing by \(\dfrac{M}{M + 100}\) separates the variables, and \(\dfrac{M + 100}{M} = 1 + \dfrac{100}{M}\) integrates to \(M + 100\ln M\). The initial condition gives the constant, and combining the logarithms gives the formula for \(t\); substituting \(M = 100\) gives 3.7396 hours, which is 3 hours 44 minutes. The towel never dries completely because the logarithm term grows without limit as \(M\) approaches 0; common errors are not converting decimal hours to minutes and integrating \(\dfrac{M + 100}{M}\) as a logarithm of \(M + 100\).

Mark scheme

1 mark
M1 Separates the variables and integrates: \(\displaystyle\int \dfrac{M + 100}{M}\,dM = \int -150\,dt\), with \(\dfrac{M + 100}{M} = 1 + \dfrac{100}{M}\) integrated to \(M + 100\ln M\) (condone sign slips). An answer with no working scores 0 marks for the show that and for the time.
An answer with no working scores 0 marks for the show that. The question demands all stages of working, so a time of 3 hours 44 minutes with no working also scores 0 marks: the substitution of \(M = 100\) into the given result (or their equation) must be seen. The explanation is marked on its own.
1 mark
A1 \(M + 100\ln M = -150t + c\) oe
1 mark
M1 (dM1) Uses \(t = 0\), \(M = 500\) to find the constant: \(c = 500 + 100\ln 500\)
Depends on the first M mark (integrating), not on the A mark just before.
1 mark
A1* \(150t = 500 - M + 100\ln 500 - 100\ln M\), and \(100\ln 500 - 100\ln M = 100\ln\dfrac{500}{M}\), giving the result cso
The combination of the logarithms must be shown before the given answer.
1 mark
M1 Substitutes \(M = 100\): \(t = \dfrac{1}{150}(400 + 100\ln 5)\)
May use the given result even if it was not shown.
1 mark
A1 3 hours 44 minutes
From \(t = 3.7396...\) hours. Accept 3 h 44 min or 224 minutes; awrt 3.74 hours (i.e. in [3.735, 3.745)) without conversion to hours and minutes is A0.
1 mark
B1 Explains that \(M = 0\) is never reached: as \(M \to 0\), \(\ln\dfrac{500}{M} \to \infty\), so \(t \to \infty\) (or \(\ln 0\) is undefined, so there is no time at which \(M = 0\)); the mass of water gets closer and closer to 0 but never reaches it
Needs a reason based on the solution (the logarithm term) or on the differential equation (e.g. \(\dfrac{dM}{dt} \to 0\) as \(M \to 0\), so the drying slows down and \(M\) only approaches 0). "It is only a model" or "because it is exponential" is B0.

Question 4

10 marks

In this question you must show all stages of your working.

A dry sponge is placed in a tray of water. The mass of water absorbed by the sponge is \(W\) grams at time \(t\) seconds after it is placed in the tray. In a model,
\[\dfrac{dW}{dt} = k\mathrm{e}^{-\frac{W}{40}}, \quad t \geqslant 0\]
where \(k\) is a positive constant.

Show that
\[W = 40\ln\left(1 + \dfrac{kt}{40}\right)\]
Given that the sponge has absorbed 40 grams of water after 5 seconds, find the exact value of \(k\).
Find, according to the model, the exact time at which the sponge has absorbed 80 grams of water. Give your answer in its simplest form.
The sponge can hold at most 120 grams of water. Explain why the model is not suitable for large values of \(t\).

Model answer

\(\displaystyle\int \mathrm{e}^{\frac{W}{40}}\,dW = \int k\,dt\)
\(40\mathrm{e}^{\frac{W}{40}} = kt + c\)
The sponge is dry at the start, so \(W = 0\) when \(t = 0\): \(40 = c\)
\(40\mathrm{e}^{\frac{W}{40}} = kt + 40\), so \(\mathrm{e}^{\frac{W}{40}} = 1 + \dfrac{kt}{40}\)
\(W = 40\ln\left(1 + \dfrac{kt}{40}\right)\), as required.
\(t = 5\), \(W = 40\): \(\ln\left(1 + \dfrac{k}{8}\right) = 1\), so \(1 + \dfrac{k}{8} = \mathrm{e}\) and \(k = 8(\mathrm{e} - 1)\)
\(W = 80\): \(\ln\left(1 + \dfrac{(\mathrm{e} - 1)t}{5}\right) = 2\), so \(\dfrac{(\mathrm{e} - 1)t}{5} = \mathrm{e}^2 - 1\)
\(t = \dfrac{5(\mathrm{e}^2 - 1)}{\mathrm{e} - 1} = \dfrac{5(\mathrm{e} - 1)(\mathrm{e} + 1)}{\mathrm{e} - 1} = 5(\mathrm{e} + 1)\) seconds
The model predicts that \(W\) keeps increasing without limit, since \(\ln\left(1 + \dfrac{kt}{40}\right) \to \infty\) as \(t \to \infty\). So for large \(t\) it predicts more than 120 grams, which the sponge cannot hold: the model is only suitable until the sponge is nearly full.

Explanation

Dividing by \(\mathrm{e}^{-\frac{W}{40}}\) gives \(\mathrm{e}^{\frac{W}{40}}\) on the left, which integrates to \(40\mathrm{e}^{\frac{W}{40}}\); the dry sponge gives \(c = 40\), and taking logarithms gives the printed form. The data give \(1 + \dfrac{k}{8} = \mathrm{e}\), and for 80 grams the factorisation \(\mathrm{e}^2 - 1 = (\mathrm{e} - 1)(\mathrm{e} + 1)\) simplifies the time to \(5(\mathrm{e} + 1)\) seconds. Common errors are multiplying instead of dividing by 40 when integrating and leaving the time unsimplified; the model fails for large \(t\) because \(W\) has no upper limit.

Mark scheme

1 mark
B1 Separates the variables correctly: \(\displaystyle\int \mathrm{e}^{\frac{W}{40}}\,dW = \int k\,dt\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the first result is a show that and the question demands working, so \(k = 8(\mathrm{e} - 1)\) and \(5(\mathrm{e} + 1)\) seconds with no working score nothing. The explanation (the last mark) is marked on its own.
1 mark
M1 Integrates both sides: \(\alpha\mathrm{e}^{\frac{W}{40}} = kt\) (\(+ c\))
1 mark
A1 \(40\mathrm{e}^{\frac{W}{40}} = kt + c\)
A common error is \(\dfrac{1}{40}\mathrm{e}^{\frac{W}{40}}\), from multiplying by \(\dfrac{1}{40}\) instead of dividing: A0.
1 mark
M1 (dM1) Uses \(W = 0\) when \(t = 0\) (the sponge is dry) to find the constant: \(c = 40\)
Depends on the first M mark (integrating), not on the A mark just before.
1 mark
A1* \(\mathrm{e}^{\frac{W}{40}} = 1 + \dfrac{kt}{40}\), so \(W = 40\ln\left(1 + \dfrac{kt}{40}\right)\) cso
A correct line with \(\mathrm{e}^{\frac{W}{40}}\) as the subject (or equivalent) must be seen before the given answer.
1 mark
M1 Substitutes \(t = 5\), \(W = 40\) and removes the logarithm: \(\ln\left(1 + \dfrac{k}{8}\right) = 1\), so \(1 + \dfrac{k}{8} = \mathrm{e}\)
May use the given result even if it was not shown.
1 mark
A1 \(k = 8(\mathrm{e} - 1)\) oe exact
Accept \(8\mathrm{e} - 8\). A decimal only (13.7) is A0.
1 mark
M1 Sets \(W = 80\) with their \(k\) and removes the logarithm: \(1 + \dfrac{kt}{40} = \mathrm{e}^2\), so \(t = \dfrac{40(\mathrm{e}^2 - 1)}{k}\)
With their exact or decimal \(k\) (a decimal \(k\) cannot lead to the exact answer, so the A mark is then lost).
1 mark
A1 \(t = 5(\mathrm{e} + 1)\) (seconds) cso
Requires the factorisation \(\mathrm{e}^2 - 1 = (\mathrm{e} - 1)(\mathrm{e} + 1)\) to simplify \(\dfrac{5(\mathrm{e}^2 - 1)}{\mathrm{e} - 1}\). Accept \(5\mathrm{e} + 5\). \(\dfrac{5(\mathrm{e}^2 - 1)}{\mathrm{e} - 1}\) is not in its simplest form: A0. A decimal only (18.6) is A0. Units are not required, but penalise wrong units.
1 mark
B1 The model predicts that \(W\) increases without limit (\(\ln\left(1 + \dfrac{kt}{40}\right) \to \infty\) as \(t \to \infty\)), so for large \(t\) it predicts more than 120 grams, which the sponge cannot hold
Needs the idea that the model's \(W\) keeps increasing with no upper limit and so would exceed 120 g (it does so from about 55.5 seconds). "It is only a model" is B0.

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