Practise Constructing differential equations. 6 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Build a paper on this topicWritten for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A radioactive substance decays so that its mass, \(m\) grams, at time \(t\) days decreases at a rate that is proportional to the mass remaining.
(a) Write down a differential equation relating \(m\) and \(t\).[2]
(b) At the instant when \(m = 50\), the mass is decreasing at a rate of 5 grams per day. Find the value of \(k\).[2]
Show the answer and mark scheme
(a) Answer: \(\frac{\mathrm{d}m}{\mathrm{d}t} = -km\), where \(k\) is a positive constant
- M1 for \(\frac{\mathrm{d}m}{\mathrm{d}t}\) proportional to the correct expression
- A1 for \(\frac{\mathrm{d}m}{\mathrm{d}t} = -km\), with the correct sign and \(k\) a positive constant
Worked solution: \(\frac{\mathrm{d}m}{\mathrm{d}t} = -km\), where \(k \gt 0\) is a constant of proportionality (the negative sign shows that the mass is decreasing).
(b) Answer: \(k = \frac{1}{10}\)
- M1 for substituting \(\frac{\mathrm{d}m}{\mathrm{d}t} = -5\) and \(m = 50\) into their equation
- A1 for \(k = \frac{1}{10}\)
Worked solution: \(-5 = -k \times 50\), so \(k = \frac{1}{10}\).
Question 2Medium7 marks
In a boarding school of 2500 students, \(I\) students are infected with a virus at time \(t\) days. A model for the spread of the virus is
\[\frac{\mathrm{d}I}{\mathrm{d}t} = 0.0004I(2500 - I) - 0.2I\]
(a) Explain, in the context of the model, what each of the terms \(0.0004I(2500 - I)\) and \(0.2I\) represents.[2]
(b) Find the non-zero value of \(I\) for which the number of infected students remains constant.[2]
(c) Show that the model predicts that the number of infected students increases whenever \(0 \lt I \lt 2000\).[2]
(d) Give one criticism of the model.[1]
Show the answer and mark scheme
(a) Answer: The first term is the rate of new infections, proportional to (infected) × (not infected); the second is the rate of recovery, proportional to the number infected.
- B1 for \(0.0004I(2500 - I)\): the rate of new infections, proportional to the product of the numbers of infected and uninfected students (the number of contacts between them)
- B1 for \(0.2I\): the rate at which infected students recover, proportional to the number infected
Worked solution: \(2500 - I\) students are not infected, so \(I(2500 - I)\) measures the number of possible contacts between infected and uninfected students: new infections occur at a rate proportional to it. Infected students recover at a rate proportional to \(I\) (on average a fraction 0.2 of them each day), which reduces \(I\).
(b) Answer: \(I = 2000\)
- M1 for \(0.0004(2500 - I) = 0.2\) (dividing by \(I \ne 0\))
- A1 for \(I = 2000\)
Worked solution: \(\frac{\mathrm{d}I}{\mathrm{d}t} = 0\) with \(I \ne 0\) gives \(0.0004(2500 - I) = 0.2\), so \(2500 - I = 500\) and \(I = 2000\).
(c) Answer: Shown
- M1 for \(\frac{\mathrm{d}I}{\mathrm{d}t} = 0.0004I(2500 - I - 500) = 0.0004I(2000 - I)\)
- A1* for both factors positive when \(0 \lt I \lt 2000\), so \(\frac{\mathrm{d}I}{\mathrm{d}t} \gt 0\) (cso)
Worked solution: \(\frac{\mathrm{d}I}{\mathrm{d}t} = 0.0004I(2500 - I) - 0.0004 \times 500I = 0.0004I(2000 - I)\). For \(0 \lt I \lt 2000\) both \(I\) and \(2000 - I\) are positive, so \(\frac{\mathrm{d}I}{\mathrm{d}t} \gt 0\) and the number infected increases.
(d) Answer: e.g. recovered students are treated as able to be infected again.
- B1 for a sensible criticism, e.g. students who recover are likely to be immune, but the model treats them as able to catch the virus again; infected students may be isolated; \(I\) should be a whole number; the school population may change
Worked solution: The model returns recovered students to the uninfected group, so they can catch the virus again; in reality most would be immune. It also predicts that 2000 students stay infected indefinitely, which is unrealistic.
Question 3Hard13 marks
Two chemicals, \(A\) and \(B\), react to form a compound \(C\). Each gram of \(C\) is formed from \(\frac{1}{3}\) gram of \(A\) and \(\frac{2}{3}\) gram of \(B\). Initially there are 4 grams of \(A\), 6 grams of \(B\) and no \(C\).
At time \(t\) minutes, \(x\) grams of \(C\) have formed. The rate at which \(C\) forms is proportional to the product of the masses of \(A\) and \(B\) that remain.
(a) Show that \(\frac{\mathrm{d}x}{\mathrm{d}t} = \lambda(12 - x)(9 - x)\), where \(\lambda\) is a positive constant.[3]
(b) State, with a reason, the greatest mass of \(C\) that can be formed.[1]
(c) Solve the differential equation to show that
\[t = \frac{1}{3\lambda}\ln\left(\frac{3(12 - x)}{4(9 - x)}\right)\][5]
(d) Given that \(x = 3\) when \(t = 10\), find the exact value of \(\lambda\).[2]
(e) Find the mass of \(C\) formed after 60 minutes, giving your answer to 3 significant figures.[2]
Show the answer and mark scheme
(a) Answer: Shown
- B1 for the remaining masses \(4 - \frac{x}{3}\) of \(A\) and \(6 - \frac{2x}{3}\) of \(B\)
- M1 for \(\frac{\mathrm{d}x}{\mathrm{d}t} = k\left(4 - \frac{x}{3}\right)\left(6 - \frac{2x}{3}\right)\)
- A1* for \(\frac{2k}{9}(12 - x)(9 - x)\), so \(\lambda = \frac{2k}{9}\) (cso)
Worked solution: When \(x\) grams of \(C\) have formed, \(\frac{x}{3}\) g of \(A\) and \(\frac{2x}{3}\) g of \(B\) have been used, leaving \(4 - \frac{x}{3} = \frac{12 - x}{3}\) g and \(6 - \frac{2x}{3} = \frac{2(9 - x)}{3}\) g.
So \(\frac{\mathrm{d}x}{\mathrm{d}t} = k \times \frac{12 - x}{3} \times \frac{2(9 - x)}{3} = \frac{2k}{9}(12 - x)(9 - x) = \lambda(12 - x)(9 - x)\), with \(\lambda = \frac{2k}{9} \gt 0\).
(b) Answer: 9 grams: \(B\) runs out when \(x = 9\).
- B1 for 9 grams, because the \(B\) is used up when \(x = 9\) (before the \(A\) is used up at \(x = 12\))
Worked solution: The \(B\) runs out when \(6 - \frac{2x}{3} = 0\), i.e. \(x = 9\), while \(A\) would last until \(x = 12\). So at most 9 g of \(C\) can form (and \(\frac{\mathrm{d}x}{\mathrm{d}t} \to 0\) as \(x \to 9\)).
(c) Answer: Shown
- B1 for separating: \(\int \frac{1}{(12 - x)(9 - x)}\,\mathrm{d}x = \int \lambda\,\mathrm{d}t\)
- M1 for a correct method for partial fractions
- A1 for \(\frac{1}{3}\left(\frac{1}{9 - x} - \frac{1}{12 - x}\right)\)
- M1 for integrating to \(\frac{1}{3}\left(\ln(12 - x) - \ln(9 - x)\right) = \lambda t + c\) and using \(x = 0\) when \(t = 0\)
- A1* for the given result (cso)
Worked solution: \(\frac{1}{(12 - x)(9 - x)} = \frac{1}{3}\left(\frac{1}{9 - x} - \frac{1}{12 - x}\right)\), so \(\frac{1}{3}\left(-\ln(9 - x) + \ln(12 - x)\right) = \lambda t + c\).
When \(t = 0\), \(x = 0\): \(c = \frac{1}{3}\ln\frac{12}{9} = \frac{1}{3}\ln\frac{4}{3}\). So \(3\lambda t = \ln\left(\frac{12 - x}{9 - x}\right) - \ln\frac{4}{3} = \ln\left(\frac{3(12 - x)}{4(9 - x)}\right)\), which gives the result.
(d) Answer: \(\lambda = \frac{1}{30}\ln\frac{9}{8}\)
- M1 for substituting: \(10 = \frac{1}{3\lambda}\ln\left(\frac{27}{24}\right)\)
- A1 for \(\lambda = \frac{1}{30}\ln\frac{9}{8}\)
Worked solution: \(\frac{3 \times 9}{4 \times 6} = \frac{9}{8}\), so \(10 = \frac{1}{3\lambda}\ln\frac{9}{8}\) and \(\lambda = \frac{1}{30}\ln\frac{9}{8}\).
(e) Answer: \(7.24\) g
- M1 for \(\frac{3(12 - x)}{4(9 - x)} = \left(\frac{9}{8}\right)^{6}\) (from \(3\lambda \times 60 = 6\ln\frac{9}{8}\)) and solving for \(x\)
- A1 for awrt 7.24 g
Worked solution: \(3\lambda \times 60 = 6\ln\frac{9}{8}\), so \(\frac{3(12 - x)}{4(9 - x)} = \left(\frac{9}{8}\right)^{6} = 2.0273\ldots\).
Then \(36 - 3x = 8.109\ldots(9 - x)\), giving \(x = \frac{36 \times (\frac{9}{8})^{6} - 36}{4 \times (\frac{9}{8})^{6} - 3} \approx 7.24\) g (3 s.f.).