| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Systems of differential equations |
| Type | Interpret model parameters from equations |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring interpretation of a predator-prey model, extraction of parameters from coupled DEs, then solving the system using eigenvalues/eigenvectors or substitution methods. Part (a) requires careful modeling interpretation, while part (b) involves substantial algebraic manipulation typical of second-order DE solving at FM level. |
| Spec | 4.10h Coupled systems: simultaneous first order DEs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a=40\), \(b=13\) | B1 | Obtains two correct values |
| \(c=1.95\), \(d=1\) | B1 | Obtains four correct values; condone "40%" and "1%"; do not accept \(-1.95\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(13y = 0.4x - \dot{x}\) (1), giving \(y = \frac{2}{65}x - \frac{1}{13}\dot{x}\) | M1 | Differentiates one equation |
| \(\dot{y} = \frac{2}{65}\dot{x} - \frac{1}{13}\ddot{x}\); substitute into (2) | M1 | Substitutes for \(\dot{y}\) or for \(x\) and \(\dot{x}\) in the other equation to eliminate one variable |
| \(\frac{2}{65}\dot{x} - \frac{1}{13}\ddot{x} = 0.01x - 1.95\); so \(5\ddot{x} - 2\dot{x} + 0.65x = 126.75\) | A1 | Forms correct simplified second order differential equation |
| CF: \(5m^2 - 2m + 0.65 = 0\); \(m = 0.2 \pm 0.3i\) | M1 | Obtains roots of auxiliary equation |
| PI: \(x = 195\) | M1 | Uses valid method to find particular integral |
| \(\therefore x = Ae^{0.2t}\cos(0.3t) + Be^{0.2t}\sin(0.3t) + 195\) | ||
| \(y = \frac{1}{65}Ae^{0.2t}\cos(0.3t) + \frac{3}{130}Ae^{0.2t}\sin(0.3t) + \frac{1}{65}Be^{0.2t}\sin(0.3t) - \frac{3}{130}Be^{0.2t}\cos(0.3t) + 6\) | A1F | States general solution for either \(x\) or \(y\) with non-zero particular integral |
| When \(t=0\): \(x=1755\) and \(y=30\); so \(1755 = A + 195\) and \(30 = \frac{A}{65} - \frac{3B}{130} + 6\); \(\Rightarrow A = 1560, B = 0\) | M1 | Uses initial conditions to find a value for each constant |
| \(x = 1560e^{0.2t}\cos(0.3t) + 195\) | A1 | States general solutions for both \(x\) and \(y\) (CAO) |
| \(y = 24e^{0.2t}\cos(0.3t) + 36e^{0.2t}\sin(0.3t) + 6\) | A1 | Writes correct solutions for both \(x\) and \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Using calculator: When \(t = 5.38\), \(x = 0\) | M1 (3.2a) | Investigates values of \(x\) for \(t > 5\); PI by \(t = 5.38\), \(x = 0\) |
| At this time \(y \approx 108\) so there are still birds of prey. The rabbits die out first. | A1 (3.2a) | Obtains a time when \(x = 0\) or \(x < 0\), and states that the rabbits die out |
| So the conservationists' plan succeeds. | E1 (3.5a) | Obtains a positive value of \(y\) for a value of \(t\) for which \(x \leq 0\) and uses their correct answers to show that the rabbits die out first. Condone no investigation of values of \(y\) between \(t = 5\) and their \(5.38\) |
## Question 14(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a=40$, $b=13$ | B1 | Obtains two correct values |
| $c=1.95$, $d=1$ | B1 | Obtains four correct values; condone "40%" and "1%"; do not accept $-1.95$ |
---
## Question 14(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $13y = 0.4x - \dot{x}$ (1), giving $y = \frac{2}{65}x - \frac{1}{13}\dot{x}$ | M1 | Differentiates one equation |
| $\dot{y} = \frac{2}{65}\dot{x} - \frac{1}{13}\ddot{x}$; substitute into (2) | M1 | Substitutes for $\dot{y}$ or for $x$ and $\dot{x}$ in the other equation to eliminate one variable |
| $\frac{2}{65}\dot{x} - \frac{1}{13}\ddot{x} = 0.01x - 1.95$; so $5\ddot{x} - 2\dot{x} + 0.65x = 126.75$ | A1 | Forms correct simplified second order differential equation |
| CF: $5m^2 - 2m + 0.65 = 0$; $m = 0.2 \pm 0.3i$ | M1 | Obtains roots of auxiliary equation |
| PI: $x = 195$ | M1 | Uses valid method to find particular integral |
| $\therefore x = Ae^{0.2t}\cos(0.3t) + Be^{0.2t}\sin(0.3t) + 195$ | | |
| $y = \frac{1}{65}Ae^{0.2t}\cos(0.3t) + \frac{3}{130}Ae^{0.2t}\sin(0.3t) + \frac{1}{65}Be^{0.2t}\sin(0.3t) - \frac{3}{130}Be^{0.2t}\cos(0.3t) + 6$ | A1F | States general solution for either $x$ or $y$ with non-zero particular integral |
| When $t=0$: $x=1755$ and $y=30$; so $1755 = A + 195$ and $30 = \frac{A}{65} - \frac{3B}{130} + 6$; $\Rightarrow A = 1560, B = 0$ | M1 | Uses initial conditions to find a value for each constant |
| $x = 1560e^{0.2t}\cos(0.3t) + 195$ | A1 | States general solutions for both $x$ and $y$ (CAO) |
| $y = 24e^{0.2t}\cos(0.3t) + 36e^{0.2t}\sin(0.3t) + 6$ | A1 | Writes correct solutions for both $x$ and $y$ |
## Question 14(c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Using calculator: When $t = 5.38$, $x = 0$ | M1 (3.2a) | Investigates values of $x$ for $t > 5$; PI by $t = 5.38$, $x = 0$ |
| At this time $y \approx 108$ so there are still birds of prey. The rabbits die out first. | A1 (3.2a) | Obtains a time when $x = 0$ or $x < 0$, and states that the rabbits die out |
| So the conservationists' plan succeeds. | E1 (3.5a) | Obtains a positive value of $y$ for a value of $t$ for which $x \leq 0$ and uses their correct answers to show that the rabbits die out first. Condone no investigation of values of $y$ between $t = 5$ and their $5.38$ |
**Question total: 14 marks**
**Paper total: 100 marks**14 On an isolated island some rabbits have been accidently introduced.
In order to eliminate them, conservationists have introduced some birds of prey.\\
At time $t$ years $( t \geq 0 )$ there are $x$ rabbits and $y$ birds of prey.\\
At time $t = 0$ there are 1755 rabbits and 30 birds of prey.\\
When $t > 0$ it is assumed that:
\begin{itemize}
\item the rabbits will reproduce at a rate of $a \%$ per year
\item each bird of prey will kill, on average, $b$ rabbits per year
\item the death rate of the birds of prey is $c$ birds per year
\item the number of birds of prey will increase at a rate of $d \%$ of the rabbit population per year.
\end{itemize}
This system is represented by the coupled differential equations:
$$\begin{aligned}
& \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4 x - 13 y \\
& \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.01 x - 1.95
\end{aligned}$$
14
\begin{enumerate}[label=(\alph*)]
\item State the value of $a$, the value of $b$, the value of $c$ and the value of $d$\\[0pt]
[2 marks]\\
14
\item Solve the coupled differential equations to find both $x$ and $y$ in terms of $t$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q14 [14]}}