Differential equations - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Differential equations for Edexcel A-Level Maths (Paper 1). 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

The number of red squirrels on an island is modelled by the differential equation
\[\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right)\]
where \(P\) is the number of squirrels \(t\) years after the start of a conservation project.
Without solving the differential equation, find the number of squirrels for which, according to the model, the population is increasing most rapidly, and find this greatest rate of increase.

Model answer

\(\dfrac{dP}{dt} = 0.4P - 0.0008P^2\)
\(\dfrac{d}{dP}\left(0.4P - 0.0008P^2\right) = 0.4 - 0.0016P = 0 \Rightarrow P = 250\)
This is a maximum because the coefficient of \(P^2\) is negative.
Greatest rate \(= 0.4 \times 250 \times \left(1 - \dfrac{250}{500}\right) = 50\) squirrels per year

Explanation

The rate of increase \(\dfrac{dP}{dt} = 0.4P - 0.0008P^2\) is a quadratic in \(P\) with a negative \(P^2\) term, so it is greatest halfway between its roots \(P = 0\) and \(P = 500\). At \(P = 250\) the rate is \(0.4 \times 250 \times 0.5 = 50\) squirrels per year. Common errors are differentiating with respect to \(t\) and giving the long-term population, 500, instead.

Mark scheme

1 mark
M1 Treats \(\dfrac{dP}{dt} = 0.4P - 0.0008P^2\) as a quadratic function of \(P\) and finds where it is greatest: differentiates with respect to \(P\) and sets the result equal to 0, or completes the square, or uses the midpoint of the roots \(P = 0\) and \(P = 500\); or correct answers \(P = 250\) and 50 squirrels per year with no working (3 marks)
An answer of \(P = 250\) and 50 squirrels per year with no working scores 3 marks. Differentiating with respect to \(t\) instead of \(P\) is M0. A correct solution of the differential equation followed by a correct maximum also scores full marks.
1 mark
A1 \(P = 250\)
1 mark
A1 Greatest rate of increase = 50 squirrels per year
Units are not required. Giving 500 (the long-term population) is A0.

Question 2

6 marks

A curve \(C\) passes through the origin. Every point \((x, y)\) on \(C\) satisfies
\[\dfrac{dy}{dx} = \mathrm{e}^{x + 2y}\]
Find an equation for \(C\), expressing \(y\) as a function of \(x\).

Model answer

\(\dfrac{dy}{dx} = \mathrm{e}^{x}\mathrm{e}^{2y}\)
\(\displaystyle\int \mathrm{e}^{-2y}\,dy = \int \mathrm{e}^{x}\,dx\)
\(-\dfrac{1}{2}\mathrm{e}^{-2y} = \mathrm{e}^{x} + c\)
\(x = 0\), \(y = 0\): \(-\dfrac{1}{2} = 1 + c\), so \(c = -\dfrac{3}{2}\)
\(\mathrm{e}^{-2y} = -2\mathrm{e}^{x} + 3 = 3 - 2\mathrm{e}^{x}\)
\(-2y = \ln(3 - 2\mathrm{e}^{x})\)
\(y = -\dfrac{1}{2}\ln(3 - 2\mathrm{e}^{x})\) (for \(x < \ln\dfrac{3}{2}\))

Explanation

Split \(\mathrm{e}^{x + 2y}\) into \(\mathrm{e}^{x}\mathrm{e}^{2y}\) with the laws of indices, then divide by \(\mathrm{e}^{2y}\) to separate the variables, so that \(\mathrm{e}^{-2y}\) integrates to \(-\dfrac{1}{2}\mathrm{e}^{-2y}\). After finding \(c = -\dfrac{3}{2}\), take the logarithm of the whole right-hand side, \(3 - 2\mathrm{e}^{x}\). Common errors are \(-2\mathrm{e}^{-2y}\) for the integral and taking logarithms term by term.

Mark scheme

1 mark
B1 Uses \(\mathrm{e}^{x + 2y} = \mathrm{e}^{x}\mathrm{e}^{2y}\) and separates: \(\displaystyle\int \mathrm{e}^{-2y}\,dy = \int \mathrm{e}^{x}\,dx\). A correct final answer of \(y = -\dfrac{1}{2}\ln(3 - 2\mathrm{e}^{x})\) with no working scores 2 marks (special case: M1 and dM1).
An answer of \(y = -\dfrac{1}{2}\ln(3 - 2\mathrm{e}^{x})\) with no working scores 2 marks (special case: the M1 for integrating and the dM1 for finding the constant are implied by the correct answer; the B1 and the A1 marks need the separation, the integrated equation and the rearrangement to be seen). This is a multi-step problem, so a bare answer cannot score full marks.
1 mark
M1 Integrates both sides: \(k\mathrm{e}^{-2y}\) or \(\mathrm{e}^{x}\)
1 mark
A1 \(-\dfrac{1}{2}\mathrm{e}^{-2y} = \mathrm{e}^{x} + c\) oe
A common error is \(-2\mathrm{e}^{-2y}\) for the integral of \(\mathrm{e}^{-2y}\): A0.
1 mark
M1 (dM1) Uses \(x = 0\), \(y = 0\) to find the constant: \(c = -\dfrac{3}{2}\)
Depends on the M mark for integrating.
1 mark
A1 \(\mathrm{e}^{-2y} = 3 - 2\mathrm{e}^{x}\) oe
e.g. \(-\dfrac{1}{2}\mathrm{e}^{-2y} = \mathrm{e}^{x} - \dfrac{3}{2}\).
1 mark
A1 \(y = -\dfrac{1}{2}\ln(3 - 2\mathrm{e}^{x})\) oe
Accept \(y = \ln\dfrac{1}{\sqrt{3 - 2\mathrm{e}^{x}}}\) or \(y = \ln\left((3 - 2\mathrm{e}^{x})^{-\frac{1}{2}}\right)\). The domain \(x < \ln\dfrac{3}{2}\) is not required. Taking logarithms term by term, e.g. \(-2y = \ln 3 - \ln 2 - x\), is A0.

Question 3

7 marks

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

During a dry spell, a garden pond slowly loses water. In a model, the pond is \(h\) metres deep, \(t\) weeks after the dry spell starts, where
\[\dfrac{dh}{dt} = -\dfrac{ht}{50}, \quad t \geqslant 0\]
When the dry spell starts, the pond is 1.2 m deep.
Find, according to the model, an expression for \(h\) in terms of \(t\).
Hence find how long it takes for the pond to become half as deep as it was at the start. Give your answer in weeks, correct to 3 significant figures.

Model answer

\(\displaystyle\int \dfrac{1}{h}\,dh = \int -\dfrac{t}{50}\,dt\)
\(\ln h = -\dfrac{t^2}{100} + c\)
\(t = 0\), \(h = 1.2\): \(c = \ln 1.2\)
\(h = 1.2\mathrm{e}^{-\frac{t^2}{100}}\)
When \(h = 0.6\): \(\mathrm{e}^{-\frac{t^2}{100}} = \dfrac{1}{2}\), so \(t^2 = 100\ln 2\)
\(t = 10\sqrt{\ln 2} = 8.33\) weeks (3 s.f.)

Explanation

Divide by \(h\) to separate the variables; the right-hand side integrates to \(-\dfrac{t^2}{100}\). The initial depth gives \(h = 1.2\mathrm{e}^{-\frac{t^2}{100}}\), and halving the depth means \(\mathrm{e}^{-\frac{t^2}{100}} = \dfrac{1}{2}\), so \(t = 10\sqrt{\ln 2} \approx 8.33\) weeks. Common errors are integrating \(\dfrac{t}{50}\) to \(\dfrac{t^2}{50}\) and taking logarithms before the exponential term is on its own.

Mark scheme

1 mark
B1 Separates the variables: \(\displaystyle\int \dfrac{1}{h}\,dh = \int -\dfrac{t}{50}\,dt\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands working, so \(h = 1.2\mathrm{e}^{-\frac{t^2}{100}}\) and 8.33 weeks with no working score nothing.
1 mark
M1 Integrates: \(\ln h\) or \(kt^2\)
1 mark
A1 \(\ln h = -\dfrac{t^2}{100} + c\)
1 mark
M1 (dM1) Uses \(t = 0\), \(h = 1.2\) to find the constant: \(c = \ln 1.2\)
Depends on the M mark for integrating. The constant should be kept exact.
1 mark
A1 \(h = 1.2\mathrm{e}^{-\frac{t^2}{100}}\) oe
Accept \(h = \mathrm{e}^{\ln 1.2 - \frac{t^2}{100}}\).
1 mark
M1 Sets \(h = 0.6\) in their expression and uses logarithms to solve for \(t\): \(-\dfrac{t^2}{100} = \ln\dfrac{1}{2}\)
Exact working until the final answer is expected.
1 mark
A1 \(t = 10\sqrt{\ln 2} =\) awrt 8.33 weeks, i.e. [8.325, 8.335)
Units are not required, but penalise wrong units. Do not accept the negative root.

Question 4

10 marks

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

A tray of cooked rice is placed in a blast chiller in which the air is kept at a constant temperature of \(-20\) °C. The temperature of the rice, \(\theta\) °C, \(t\) minutes after it is placed in the chiller, is modelled by the differential equation
\[\dfrac{d\theta}{dt} = -k(\theta + 20)\]
where \(k\) is a positive constant.
The temperature of the rice is 90 °C when it is placed in the chiller and 50 °C after 15 minutes.
Food safety guidance says that the rice should be cooled to 3 °C or below within 90 minutes of being placed in the chiller.
Use the model to determine whether the rice meets this guidance.
Give one limitation of the model.

Model answer

\(\displaystyle\int \dfrac{1}{\theta + 20}\,d\theta = \int -k\,dt\)
\(\ln(\theta + 20) = -kt + c\)
\(t = 0\), \(\theta = 90\): \(c = \ln 110\), so \(\theta + 20 = 110\mathrm{e}^{-kt}\)
\(\theta = 110\mathrm{e}^{-kt} - 20\)
\(t = 15\), \(\theta = 50\): \(70 = 110\mathrm{e}^{-15k}\), so \(k = \dfrac{1}{15}\ln\dfrac{11}{7} = 0.0301\) (3 s.f.)
\(\theta = 3\): \(23 = 110\mathrm{e}^{-kt}\), so \(t = \dfrac{1}{k}\ln\dfrac{110}{23} = 51.9\) minutes (3 s.f.)
51.9 minutes is less than 90 minutes, so according to the model the rice meets the guidance.
Limitation: the air temperature in the chiller will rise when the hot rice is put in, so it will not stay constant at \(-20\) °C.

Explanation

Separate and integrate to get \(\theta = 110\mathrm{e}^{-kt} - 20\), then use the second reading to find \(k = \dfrac{1}{15}\ln\dfrac{11}{7}\). Setting \(\theta = 3\) gives about 51.9 minutes, comfortably within 90 minutes. Common errors are using 90 as the constant of integration instead of 110, and limitations that are not about this context.

Mark scheme

1 mark
B1 Separates the variables: \(\displaystyle\int \dfrac{1}{\theta + 20}\,d\theta = \int -k\,dt\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands working, so a time of about 52 minutes, or "yes", with no working scores nothing. The limitation (the last mark) is marked on its own.
1 mark
M1 Integrates: \(\ln(\theta + 20)\) or \(-kt\)
1 mark
A1 \(\ln(\theta + 20) = -kt + c\)
Modulus signs are not needed, since \(\theta + 20 > 0\).
1 mark
M1 (dM1) Uses \(t = 0\), \(\theta = 90\) to find the constant: \(c = \ln 110\)
Depends on the M mark for integrating.
1 mark
A1 \(\theta = 110\mathrm{e}^{-kt} - 20\) oe
1 mark
M1 Uses \(t = 15\), \(\theta = 50\) to form \(70 = 110\mathrm{e}^{-15k}\) (or equivalent) and solves for \(k\) using logarithms
1 mark
A1 \(k = \dfrac{1}{15}\ln\dfrac{11}{7} =\) awrt 0.0301, i.e. [0.03005, 0.03015)
1 mark
M1 Sets \(\theta = 3\) in their equation and solves for \(t\): \(\mathrm{e}^{-kt} = \dfrac{23}{110}\)
Alternative method: finds the temperature after 90 minutes, \(110\mathrm{e}^{-90k} - 20\).
1 mark
A1 \(t =\) awrt 51.9 minutes, i.e. [51.85, 51.95) (or awrt 52.0, i.e. [51.95, 52.05), from \(k = 0.0301\)), which is less than 90 minutes, so the rice meets the guidance
Requires the time and a correct conclusion. By the alternative method: the temperature after 90 minutes is awrt \(-12.7\) °C (i.e. [−12.75, −12.65)), which is below 3 °C, so the rice meets the guidance.
1 mark
B1 A limitation in context, e.g. the air in the chiller will not stay at exactly \(-20\) °C (it warms when the hot rice is put in); the rice does not cool evenly (the middle stays warmer than the surface); the model predicts that the rice cools towards \(-20\) °C, but once it freezes the rate of cooling changes
Do not accept vague or non-contextual comments such as "it is only a model", "the temperature is not exact" or "other factors".

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