| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.8 This is a separable differential equation requiring partial fractions, integration of trigonometric functions, and manipulation to reach a specific form. While the individual techniques are C4 standard, the combination of steps (partial fractions, separating variables, integrating cos 2t, applying initial conditions, and algebraic manipulation to the given form) makes this more demanding than a typical textbook exercise. Part (c) adds further algebraic complexity with logarithms and inverse trig functions. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| \(\frac{2}{P(P-2)} = \frac{A}{P} + \frac{B}{(P-2)}\) | |
| \(2 = A(P-2) + BP\) | M1 |
| Can be implied. | |
| \(A = -1\), \(B = 1\) | A1 |
| Either one. | |
| \(\frac{1}{(P-2)} - \frac{1}{P}\) or any equivalent form. | A1 |
| This answer cannot be recovered from part (b). | |
| [3] |
| Answer | Marks |
|---|---|
| \(\frac{dP}{dt} = \frac{1}{2}P(P-2)\cos 2t\) | |
| \(\int\frac{P(P-2)}{dP} = \int\cos 2t \, dt\) | B1 oe |
| can be implied by later working | |
| \(\ln(P-2) - \ln P = \frac{1}{2}\sin 2t\) \((+c)\) | M1 |
| \(\pm \lambda\ln(P-\beta) \pm \mu\ln P = \pm K\sin\delta t + c\), \(\lambda, \mu, \beta, \delta \neq 0\), applies a fully correct method to eliminate their logarithms. Must have a constant of integration that need not be evaluated (see note) | M1 |
| \(\{t = 0, P = 3 \Rightarrow \} \ln1 - \ln3 = 0 + c\) | M1 |
| \(\Rightarrow c = -\ln3\) or \(\ln(t)\) | |
| \(\ln(P-2) - \ln P = \frac{1}{2}\sin 2t - \ln3\) | |
| \(\ln\left(\frac{3(P-2)}{P}\right) = \frac{1}{2}\sin 2t\) | |
| \(\frac{3(P-2)}{P} = e^{\frac{1}{2}\sin 2t}\) | |
| \(3(P-2) = Pe^{\frac{1}{2}\sin 2t} \Rightarrow 3P - 6 = Pe^{\frac{1}{2}\sin 2t}\) | |
| gives \(3P - Pe^{\frac{1}{2}\sin 2t} = 6 \Rightarrow P(3-e^{\frac{1}{2}\sin 2t}) = 6\) | dM1 |
| \(P = \frac{6}{(3-e^{\frac{1}{2}\sin 2t})}\) | A1 * cso |
| [7] |
| Answer | Marks |
|---|---|
| [Population \(= 4000 \Rightarrow\)] \(P = 4\) or applies \(P = 4\) | M1 |
| Obtains \(\pm \lambda\sin 2t = \ln k\) or \(\pm \lambda\sin t = \ln k\), where \(\lambda\) and \(k\) are numerical values and \(\lambda\) can be 1 | M1 |
| \(t = 0.4728700467\ldots\) | A1 |
| anything that rounds to 0.473. Do not apply isw here. | |
| [3] |
## Part (a)
$\frac{2}{P(P-2)} = \frac{A}{P} + \frac{B}{(P-2)}$ | |
$2 = A(P-2) + BP$ | M1 |
Can be implied. | |
$A = -1$, $B = 1$ | A1 |
Either one. | |
$\frac{1}{(P-2)} - \frac{1}{P}$ or any equivalent form. | A1 |
This answer **cannot** be recovered from part (b). | |
| [3] |
## Part (b)
$\frac{dP}{dt} = \frac{1}{2}P(P-2)\cos 2t$ | |
$\int\frac{P(P-2)}{dP} = \int\cos 2t \, dt$ | B1 oe |
can be implied by later working | |
$\ln(P-2) - \ln P = \frac{1}{2}\sin 2t$ $(+c)$ | M1 |
$\pm \lambda\ln(P-\beta) \pm \mu\ln P = \pm K\sin\delta t + c$, $\lambda, \mu, \beta, \delta \neq 0$, applies a fully correct method to eliminate their logarithms. Must have a constant of integration that need not be evaluated (see note) | M1 |
$\{t = 0, P = 3 \Rightarrow \} \ln1 - \ln3 = 0 + c$ | M1 |
$\Rightarrow c = -\ln3$ or $\ln(t)$ | |
$\ln(P-2) - \ln P = \frac{1}{2}\sin 2t - \ln3$ | |
$\ln\left(\frac{3(P-2)}{P}\right) = \frac{1}{2}\sin 2t$ | |
$\frac{3(P-2)}{P} = e^{\frac{1}{2}\sin 2t}$ | |
$3(P-2) = Pe^{\frac{1}{2}\sin 2t} \Rightarrow 3P - 6 = Pe^{\frac{1}{2}\sin 2t}$ | |
gives $3P - Pe^{\frac{1}{2}\sin 2t} = 6 \Rightarrow P(3-e^{\frac{1}{2}\sin 2t}) = 6$ | dM1 |
$P = \frac{6}{(3-e^{\frac{1}{2}\sin 2t})}$ | A1 * cso |
| [7] |
## Part (c)
[Population $= 4000 \Rightarrow$] $P = 4$ or applies $P = 4$ | M1 |
Obtains $\pm \lambda\sin 2t = \ln k$ or $\pm \lambda\sin t = \ln k$, where $\lambda$ and $k$ are numerical values and $\lambda$ can be 1 | M1 |
$t = 0.4728700467\ldots$ | A1 |
anything that rounds to 0.473. Do not apply isw here. | |
| [3] |
---\begin{enumerate}[label=(\alph*)]
\item Express $\frac{2}{P(P-2)}$ in partial fractions.
[3]
\end{enumerate}
A team of biologists is studying a population of a particular species of animal.
The population is modelled by the differential equation
$$\frac{dP}{dt} = \frac{1}{2}P(P-2)\cos 2t, \quad t \geqslant 0$$
where $P$ is the population in thousands, and $t$ is the time measured in years since the start of the study.
Given that $P = 3$ when $t = 0$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item solve this differential equation to show that
$$P = \frac{6}{3 - e^{\frac{1}{2}\sin 2t}}$$
[7]
\item find the time taken for the population to reach 4000 for the first time.
Give your answer in years to 3 significant figures.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2015 Q7 [13]}}