| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Modelling with logistic growth |
| Difficulty | Standard +0.8 This is a standard logistic growth differential equation requiring partial fractions, separation of variables, and integration with an integrating factor approach. While it involves multiple steps (partial fractions, integration, applying initial conditions, and algebraic manipulation), each technique is routine for Further Maths students. The question is well-scaffolded with part (a) giving the partial fractions setup, making it more accessible than if students had to identify this approach independently. It's moderately challenging due to the length and algebraic manipulation required, but doesn't require novel insight beyond applying standard Further Maths techniques. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(\frac{1}{P(11-2P)} = \frac{A}{P} + \frac{B}{(11-2P)}\) | B1 | |
| Substitutes \(P=0\) or \(P=\frac{11}{2}\) into \(1 = A(11-2P) + BP \Rightarrow A\) or \(B\) | M1 | Alternatively compares terms to set up and solve two simultaneous equations in \(A\) and \(B\) |
| \(\frac{1}{P(11-2P)} = \frac{1/11}{P} + \frac{2/11}{(11-2P)}\) | A1 | Or equivalent e.g. \(\frac{1}{11P} + \frac{2}{11(11-2P)}\). Correct answer with no working scores all 3 marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Separates variables: \(\int \frac{22}{P(11-2P)}\,dP = \int 1\,dt\) | B1 | |
| Uses (a) and attempts to integrate: \(\int \frac{2}{P} + \frac{4}{(11-2P)}\,dP = t + c\) | M1 | Uses part (a) and \(\int \frac{A}{P} + \frac{B}{(11-2P)}dP = A\ln P \pm C\ln(11-2P)\) |
| \(2\ln P - 2\ln(11-2P) = t + c\) | A1 | Integrates both sides to form correct equation including \(c\) |
| Substitutes \(t=0, P=1 \Rightarrow c = (-2\ln 9)\) | M1 | |
| Substitutes \(P=2 \Rightarrow t = 2\ln 2 + 2\ln 9 - 2\ln 7\) | M1 | Dependent on both previous M marks |
| Time \(= 1.89\) years | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses ln laws: \(2\ln P - 2\ln(11-2P) = t - 2\ln 9 \Rightarrow \ln\left(\frac{9P}{11-2P}\right) = \frac{1}{2}t\) | M1 | Uses correct log laws to move from \(2\ln P - 2\ln(11-2P) = t+c\) to \(\ln\left(\frac{P}{11-2P}\right) = \frac{1}{2}t + d\) for their numerical \(c\) |
| Makes \(P\) the subject: \(\frac{9P}{11-2P} = e^{\frac{1}{2}t} \Rightarrow 9P = (11-2P)e^{\frac{1}{2}t}\) | M1 | Uses correct method to get \(P\) in terms of \(e^{\frac{1}{2}t}\) |
| \(\Rightarrow P = \dfrac{11}{2 + 9e^{-\frac{1}{2}t}} \Rightarrow A=11,\, B=2,\, C=9\) | A1 | Achieves correct answer in the form required |
## Question 16:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{1}{P(11-2P)} = \frac{A}{P} + \frac{B}{(11-2P)}$ | B1 | |
| Substitutes $P=0$ or $P=\frac{11}{2}$ into $1 = A(11-2P) + BP \Rightarrow A$ or $B$ | M1 | Alternatively compares terms to set up and solve two simultaneous equations in $A$ and $B$ |
| $\frac{1}{P(11-2P)} = \frac{1/11}{P} + \frac{2/11}{(11-2P)}$ | A1 | Or equivalent e.g. $\frac{1}{11P} + \frac{2}{11(11-2P)}$. Correct answer with no working scores all 3 marks. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Separates variables: $\int \frac{22}{P(11-2P)}\,dP = \int 1\,dt$ | B1 | |
| Uses (a) and attempts to integrate: $\int \frac{2}{P} + \frac{4}{(11-2P)}\,dP = t + c$ | M1 | Uses part (a) and $\int \frac{A}{P} + \frac{B}{(11-2P)}dP = A\ln P \pm C\ln(11-2P)$ |
| $2\ln P - 2\ln(11-2P) = t + c$ | A1 | Integrates both sides to form correct equation including $c$ |
| Substitutes $t=0, P=1 \Rightarrow c = (-2\ln 9)$ | M1 | |
| Substitutes $P=2 \Rightarrow t = 2\ln 2 + 2\ln 9 - 2\ln 7$ | M1 | Dependent on both previous M marks |
| Time $= 1.89$ years | A1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses ln laws: $2\ln P - 2\ln(11-2P) = t - 2\ln 9 \Rightarrow \ln\left(\frac{9P}{11-2P}\right) = \frac{1}{2}t$ | M1 | Uses correct log laws to move from $2\ln P - 2\ln(11-2P) = t+c$ to $\ln\left(\frac{P}{11-2P}\right) = \frac{1}{2}t + d$ for their numerical $c$ |
| Makes $P$ the subject: $\frac{9P}{11-2P} = e^{\frac{1}{2}t} \Rightarrow 9P = (11-2P)e^{\frac{1}{2}t}$ | M1 | Uses correct method to get $P$ in terms of $e^{\frac{1}{2}t}$ |
| $\Rightarrow P = \dfrac{11}{2 + 9e^{-\frac{1}{2}t}} \Rightarrow A=11,\, B=2,\, C=9$ | A1 | Achieves correct answer in the form required |\begin{enumerate}
\item (a) Express $\frac { 1 } { P ( 11 - 2 P ) }$ in partial fractions.
\end{enumerate}
A population of meerkats is being studied.\\
The population is modelled by the differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 22 } P ( 11 - 2 P ) , \quad t \geqslant 0 , \quad 0 < P < 5.5$$
where $P$, in thousands, is the population of meerkats and $t$ is the time measured in years since the study began.
Given that there were 1000 meerkats in the population when the study began,\\
(b) determine the time taken, in years, for this population of meerkats to double,\\
(c) show that
$$P = \frac { A } { B + C \mathrm { e } ^ { - \frac { 1 } { 2 } t } }$$
where $A , B$ and $C$ are integers to be found.
\hfill \mbox{\textit{Edexcel Paper 2 Q16 [12]}}