| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| 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 logistic differential equation requiring separation of variables with partial fractions decomposition of 1/(N(1800-N)), followed by integration and application of initial conditions. While the technique is standard for Further Maths, the algebraic manipulation and handling of logarithms to reach the final explicit form N(t) = 1800/(1+5e^(-t/2)) requires careful execution across multiple steps, making it moderately challenging but within expected FM scope. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Separate variables correctly and integrate at least one side | M1 | |
| Carry out an attempt to find \(A\) and \(B\) such that \(\frac{1}{N(1800-N)} \equiv \frac{A}{N} + \frac{B}{1800-N}\), or equivalent | M1 | |
| Obtain \(\frac{2}{N} + \frac{2}{1800-N}\) or equivalent | A1 | |
| Integrates to produce two terms involving natural logarithms | M1 | |
| Obtain \(2\ln N - 2\ln(1800-N) = t\) or equivalent | A1 | |
| Evaluate a constant, or use \(N = 300\) and \(t = 0\) in a solution involving \(a\ln N\), \(b\ln(1800)\) and \(ct\) | M1 | |
| Obtain \(2\ln N - 2\ln(1800-N) = t - 2\ln 5\) or equivalent | A1 | |
| Use laws of logarithms to remove logarithms | M1 | |
| Obtain \(N = \dfrac{1800e^{\frac{1}{2}t}}{5 + e^{\frac{1}{2}t}}\) or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply that \(N\) approaches \(1800\) | B1 |
# Question 10:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Separate variables correctly and integrate at least one side | M1 | |
| Carry out an attempt to find $A$ and $B$ such that $\frac{1}{N(1800-N)} \equiv \frac{A}{N} + \frac{B}{1800-N}$, or equivalent | M1 | |
| Obtain $\frac{2}{N} + \frac{2}{1800-N}$ or equivalent | A1 | |
| Integrates to produce two terms involving natural logarithms | M1 | |
| Obtain $2\ln N - 2\ln(1800-N) = t$ or equivalent | A1 | |
| Evaluate a constant, or use $N = 300$ and $t = 0$ in a solution involving $a\ln N$, $b\ln(1800)$ and $ct$ | M1 | |
| Obtain $2\ln N - 2\ln(1800-N) = t - 2\ln 5$ or equivalent | A1 | |
| Use laws of logarithms to remove logarithms | M1 | |
| Obtain $N = \dfrac{1800e^{\frac{1}{2}t}}{5 + e^{\frac{1}{2}t}}$ or equivalent | A1 | |
**Total: [9]**
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply that $N$ approaches $1800$ | B1 | |
**Total: [1]**10 The number of birds of a certain species in a forested region is recorded over several years. At time $t$ years, the number of birds is $N$, where $N$ is treated as a continuous variable. The variation in the number of birds is modelled by
$$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { N ( 1800 - N ) } { 3600 }$$
It is given that $N = 300$ when $t = 0$.\\
(i) Find an expression for $N$ in terms of $t$.\\
(ii) According to the model, how many birds will there be after a long time?
\hfill \mbox{\textit{CAIE P3 2011 Q10 [10]}}