| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.8 This is a logistic differential equation requiring separation of variables with partial fractions decomposition, followed by integration and algebraic manipulation to reach a specific form. While the technique is standard for C4, the partial fractions and exponential manipulation require careful execution across multiple steps, making it moderately challenging but still within typical A-level scope. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\int \frac{1}{P(5-P)}\,dP = \int \frac{1}{15}\,dt\) | B1 | |
| \(1 = A(5-P) + BP\) | M1 | |
| \(A = \frac{1}{5},\, B = \frac{1}{5}\) | A1 | |
| \(\int \frac{1}{P(5-P)}\,dP = \int \frac{\frac{1}{5}}{P} + \frac{\frac{1}{5}}{5-P}\,dP\) | A1 | |
| \(\Rightarrow \frac{1}{5}\ln P - \frac{1}{5}\ln(5-P) = \frac{1}{15}t \quad (+c)\) | M1 A1ft | |
| \(\{t=0,\,P=1\} \Rightarrow \frac{1}{5}\ln 1 - \frac{1}{5}\ln 4 = 0 + c \Rightarrow c = -\frac{1}{5}\ln 4\) | dM1 | |
| \(\frac{1}{5}\ln\!\left(\frac{P}{5-P}\right) = \frac{1}{15}t - \frac{1}{5}\ln 4\) | ||
| \(\ln\!\left(\frac{4P}{5-P}\right) = \frac{1}{3}t\) | dM1 | Using log laws correctly |
| \(\frac{4P}{5-P} = e^{\frac{1}{3}t}\) or \(\frac{5-P}{4P} = e^{-\frac{1}{3}t}\) | dM1 | Eliminate ln's correctly |
| \(4P = 5e^{\frac{1}{3}t} - Pe^{\frac{1}{3}t} \Rightarrow P(4+e^{\frac{1}{3}t}) = 5e^{\frac{1}{3}t}\) | ||
| \(P = \frac{5e^{\frac{1}{3}t}}{4+e^{\frac{1}{3}t}}\) | dM1 | Make \(P\) the subject |
| \(P = \frac{5}{1+4e^{-\frac{1}{3}t}}\) or \(P = \frac{25}{5+20e^{-\frac{1}{3}t}}\) | A1 | (11) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(1 + 4e^{-\frac{1}{3}t} > 1 \Rightarrow P < 5\). So population cannot exceed 5000 | B1 | (1), (12 marks) |
## Question 11:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\int \frac{1}{P(5-P)}\,dP = \int \frac{1}{15}\,dt$ | B1 | |
| $1 = A(5-P) + BP$ | M1 | |
| $A = \frac{1}{5},\, B = \frac{1}{5}$ | A1 | |
| $\int \frac{1}{P(5-P)}\,dP = \int \frac{\frac{1}{5}}{P} + \frac{\frac{1}{5}}{5-P}\,dP$ | A1 | |
| $\Rightarrow \frac{1}{5}\ln P - \frac{1}{5}\ln(5-P) = \frac{1}{15}t \quad (+c)$ | M1 A1ft | |
| $\{t=0,\,P=1\} \Rightarrow \frac{1}{5}\ln 1 - \frac{1}{5}\ln 4 = 0 + c \Rightarrow c = -\frac{1}{5}\ln 4$ | dM1 | |
| $\frac{1}{5}\ln\!\left(\frac{P}{5-P}\right) = \frac{1}{15}t - \frac{1}{5}\ln 4$ | | |
| $\ln\!\left(\frac{4P}{5-P}\right) = \frac{1}{3}t$ | dM1 | Using log laws correctly |
| $\frac{4P}{5-P} = e^{\frac{1}{3}t}$ or $\frac{5-P}{4P} = e^{-\frac{1}{3}t}$ | dM1 | Eliminate ln's correctly |
| $4P = 5e^{\frac{1}{3}t} - Pe^{\frac{1}{3}t} \Rightarrow P(4+e^{\frac{1}{3}t}) = 5e^{\frac{1}{3}t}$ | | |
| $P = \frac{5e^{\frac{1}{3}t}}{4+e^{\frac{1}{3}t}}$ | dM1 | Make $P$ the subject |
| $P = \frac{5}{1+4e^{-\frac{1}{3}t}}$ or $P = \frac{25}{5+20e^{-\frac{1}{3}t}}$ | A1 | (11) |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $1 + 4e^{-\frac{1}{3}t} > 1 \Rightarrow P < 5$. So population cannot exceed 5000 | B1 | (1), **(12 marks)** |11. A team of conservationists is studying the population of meerkats on a nature reserve. The population is modelled by the differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 15 } P ( 5 - P ) , \quad t \geqslant 0$$
where $P$, in thousands, is the population of meerkats and $t$ is the time measured in years since the study began.
Given that when $t = 0 , P = 1$,
\begin{enumerate}[label=(\alph*)]
\item solve the differential equation, giving your answer in the form
$$P = \frac { a } { b + c \mathrm { e } ^ { - \frac { 1 } { 3 } t } }$$
where $a$, $b$ and $c$ are integers.
\item Hence show that the population cannot exceed 5000
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 Q11 [12]}}