| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | December |
| Marks | 16 |
| Topic | Systems of differential equations |
| Type | Determine long-term or comparative behaviour |
| Difficulty | Standard +0.8 This is a Further Maths question requiring conversion of coupled first-order DEs to second-order, solving auxiliary equations, finding general and particular solutions, and interpreting long-term behavior. While systematic, it demands multiple techniques and careful algebraic manipulation across several parts, placing it moderately above average difficulty. |
| Spec | 4.10h Coupled systems: simultaneous first order DEs |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dx}{dt} = y \Rightarrow \frac{d^2x}{dt^2} = \frac{dy}{dt} = 2y - 5x\) | M1 | Differentiate 1st DE and substitute back in |
| \(\Rightarrow \frac{d^2x}{dt^2} = 2\frac{dx}{dt} - 5x\) | A1 | AG |
| (b)(i) \(\frac{d^2x}{dt^2} - 2\frac{dx}{dt} + 5x = 0\); A.E. is \(n^2 - 2n + 5 = 0\); \(\Rightarrow n = 1 \pm 2i\); \(\Rightarrow x = e^t(A\sin 2t + B\cos 2t)\) | M1, A1, A1 | For auxiliary equation; BC |
| (b)(ii) \(x = e^t(A\sin 2t + B\cos 2t)\) and \(\frac{dx}{dt} = y\); \(\Rightarrow \frac{dx}{dt} = e^t(A\sin 2t + B\cos 2t) + e^t(2A\cos 2t - 2B\sin 2t)\); \(\Rightarrow y = e^t((A-2B)\sin 2t + (2A+B)\cos 2t)\) | M1, A1 | |
| (c) \(x = e^t(A\sin 2t + B\cos 2t)\); \(x = 1, t = 0\); \(\Rightarrow 1 = B\); \(\Rightarrow y = e^t((A-2)\sin 2t + (2A+1)\cos 2t)\); \(y = 3, t = 0\); \(\Rightarrow 3 = 2A + 1 \Rightarrow A = 1\); \(\Rightarrow x = e^t(\sin 2t + \cos 2t), y = e^t(3\cos 2t - \sin 2t)\) | M1, A1, A1, A1 | Translate context into condition to find \(B\); Translate context into condition to find \(A\); Both stated with \(A\) and \(B\) substituted for |
| (d) \(x = 0 \Rightarrow x = e^t(\sin 2t + \cos 2t) = 0\); \(t = 1.178...\); So 1.12 years; \(y = 0 \Rightarrow y = e^t(3\cos 2t - \sin 2t) = 0\); \(t = 0.624...\); So 0.625 years; \(0.625 < 1.12\) so the prey die out first. | M1, A1, A1, A1, E1 | Set either their \(x = 0\) or their \(y = 0\) and solve; BC; BC |
**(a)** $\frac{dx}{dt} = y \Rightarrow \frac{d^2x}{dt^2} = \frac{dy}{dt} = 2y - 5x$ | M1 | Differentiate 1st DE and substitute back in
$\Rightarrow \frac{d^2x}{dt^2} = 2\frac{dx}{dt} - 5x$ | A1 | AG
**(b)(i)** $\frac{d^2x}{dt^2} - 2\frac{dx}{dt} + 5x = 0$; A.E. is $n^2 - 2n + 5 = 0$; $\Rightarrow n = 1 \pm 2i$; $\Rightarrow x = e^t(A\sin 2t + B\cos 2t)$ | M1, A1, A1 | For auxiliary equation; BC
**(b)(ii)** $x = e^t(A\sin 2t + B\cos 2t)$ and $\frac{dx}{dt} = y$; $\Rightarrow \frac{dx}{dt} = e^t(A\sin 2t + B\cos 2t) + e^t(2A\cos 2t - 2B\sin 2t)$; $\Rightarrow y = e^t((A-2B)\sin 2t + (2A+B)\cos 2t)$ | M1, A1 |
**(c)** $x = e^t(A\sin 2t + B\cos 2t)$; $x = 1, t = 0$; $\Rightarrow 1 = B$; $\Rightarrow y = e^t((A-2)\sin 2t + (2A+1)\cos 2t)$; $y = 3, t = 0$; $\Rightarrow 3 = 2A + 1 \Rightarrow A = 1$; $\Rightarrow x = e^t(\sin 2t + \cos 2t), y = e^t(3\cos 2t - \sin 2t)$ | M1, A1, A1, A1 | Translate context into condition to find $B$; Translate context into condition to find $A$; Both stated with $A$ and $B$ substituted for
**(d)** $x = 0 \Rightarrow x = e^t(\sin 2t + \cos 2t) = 0$; $t = 1.178...$; So 1.12 years; $y = 0 \Rightarrow y = e^t(3\cos 2t - \sin 2t) = 0$; $t = 0.624...$; So 0.625 years; $0.625 < 1.12$ so the prey die out first. | M1, A1, A1, A1, E1 | Set either their $x = 0$ or their $y = 0$ and solve; BC; BC
---10 In a predator-prey environment the population, at time $t$ years, of predators is $x$ and prey is $y$. The populations of predators and prey are measured in hundreds.
The populations are modelled by the following simultaneous differential equations.
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = y \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } = 2 y - 5 x$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 5 x$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the general solution for $x$.
\item Find the equivalent general solution for $y$.
Initially there are 100 predators and 300 prey.
\end{enumerate}\item Find the particular solutions for $x$ and $y$.
\item Determine whether the model predicts that the predators will die out before the prey.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q10 [16]}}