| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Deriving the differential equation |
| Difficulty | Standard +0.3 This is a straightforward modelling question where part (a) requires simple percentage-to-rate conversion, part (b) involves writing down a differential equation by combining given rates (no derivation needed), part (c) is a standard integrating factor application, and part (d) asks for generic model limitations. All steps are routine for Further Maths students with no novel problem-solving required. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dt} = -ky\), separates: \(\int\frac{1}{y}\,dy = \int -k\,dt\) | M1 | Separates variables (AO 1.1a) |
| \(y = y_0 e^{-kt}\) | M1 | Deduces exponential form (AO 2.2a) |
| \(0.84y_0 = y_0 e^{-k}\) | M1 | Forms correct equation (AO 3.3) |
| \(k = 0.174\) | A1 | Obtains correct answer (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dt} = -0.174y + 45 + 20t\) | M1 | Forms DE with three terms and \(\frac{dy}{dt}\) (AO 3.3) |
| Correct DE consistent with answer to part (a) | A1F | May use \(k\) in place of \(0.174\) (AO 3.3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Splits into CF and PI, or finds an integrating factor | M1 | AO 3.1a |
| CF: \(y = Ae^{-0.174t}\); PI: \(y = p + qt\), giving \(q + 0.174p + 0.174qt = 45 + 20t\) | M1 | Obtains PI of form \(p + qt\), or uses integration by parts for \(te^{kt}\) term (AO 1.1a) |
| \(q = \frac{20}{0.174} = 115\), \(p = \frac{45-q}{0.174} = -402\) | A1F | Obtains correct general solution from their value of \(k\) (AO 1.1b) |
| General solution: \(y = Ae^{-0.174t} - 402 + 115t\); \(t=0 \Rightarrow y = 340\), so \(A = 742\) | M1 | Uses initial conditions to find unknown constant for general solution that includes exponential term (AO 3.3) |
| \(y = 742e^{-0.174t} - 402 + 115t\) | A1 | Obtains correct solution (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Births would occur at a particular time of year, not at a steady rate | E1 | One reasonable limitation of model (AO 3.5b) |
| Over a long period of time the population would increase indefinitely according to the model | E1 | One reasonable limitation of model (AO 3.5b) |
# Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dt} = -ky$, separates: $\int\frac{1}{y}\,dy = \int -k\,dt$ | M1 | Separates variables (AO 1.1a) |
| $y = y_0 e^{-kt}$ | M1 | Deduces exponential form (AO 2.2a) |
| $0.84y_0 = y_0 e^{-k}$ | M1 | Forms correct equation (AO 3.3) |
| $k = 0.174$ | A1 | Obtains correct answer (AO 1.1b) |
# Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dt} = -0.174y + 45 + 20t$ | M1 | Forms DE with three terms and $\frac{dy}{dt}$ (AO 3.3) |
| Correct DE consistent with answer to part (a) | A1F | May use $k$ in place of $0.174$ (AO 3.3) |
# Question 10(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Splits into CF and PI, or finds an integrating factor | M1 | AO 3.1a |
| CF: $y = Ae^{-0.174t}$; PI: $y = p + qt$, giving $q + 0.174p + 0.174qt = 45 + 20t$ | M1 | Obtains PI of form $p + qt$, or uses integration by parts for $te^{kt}$ term (AO 1.1a) |
| $q = \frac{20}{0.174} = 115$, $p = \frac{45-q}{0.174} = -402$ | A1F | Obtains correct general solution from their value of $k$ (AO 1.1b) |
| General solution: $y = Ae^{-0.174t} - 402 + 115t$; $t=0 \Rightarrow y = 340$, so $A = 742$ | M1 | Uses initial conditions to find unknown constant for general solution that includes exponential term (AO 3.3) |
| $y = 742e^{-0.174t} - 402 + 115t$ | A1 | Obtains correct solution (AO 1.1b) |
# Question 10(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Births would occur at a particular time of year, not at a steady rate | E1 | One reasonable limitation of model (AO 3.5b) |
| Over a long period of time the population would increase indefinitely according to the model | E1 | One reasonable limitation of model (AO 3.5b) |10 In a colony of seabirds, there are $y$ birds at time $t$ years.
10
\begin{enumerate}[label=(\alph*)]
\item The rate of reduction in the number of birds due to birds dying or leaving the colony is proportional to the number of birds.
In one year the reduction in the number of birds due to birds dying or leaving the colony is equal to $16 \%$ of the number of birds at the start of the year.
If no birds are born or join the colony, find the constant $k$ such that
$$\frac { \mathrm { d } y } { \mathrm {~d} t } = - k y$$
Give your answer to three significant figures.\\
10
\item A wildlife protection group takes measures to support the colony.\\
The rate of reduction in the number of birds due to birds dying or leaving the colony is the same as in part (a), but in addition:
\begin{itemize}
\item The rate of increase in the number of birds due to births is $20 t$ per year.
\item The wildlife protection group brings 45 birds into the colony each year.
\end{itemize}
Write down a first-order differential equation for $y$ and $t$\\
10
\item The initial number of birds is 340
Solve your differential equation from part (b) to find $y$ in terms of $t$\\
10
\item Describe two limitations of the model you have used.
Limitation 1 $\_\_\_\_$\\
Limitation 2 $\_\_\_\_$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2021 Q10 [13]}}