| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Deriving the differential equation |
| Difficulty | Standard +0.8 This is a moderately challenging question requiring students to formulate a differential equation from a word problem involving proportionality, then use the given initial condition to find the constant of proportionality. Part (a) requires careful algebraic manipulation to derive the specific form shown, while part (b) involves standard but multi-step techniques (partial fractions, separation of variables, logarithm manipulation). The modeling aspect and the need to work backwards from given information elevates this above routine exercises. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply equation of the form \(\dfrac{dx}{dt} = kx(300-x)\) and use \(\dfrac{dx}{dt} = 0.2\) and \(x=1\) | M1 | M0 for verification |
| Obtain \(k = \dfrac{1}{1495}\) and rearrange to given answer | A1 | \(1495\dfrac{dx}{dt} = x(300-x)\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly | B1 | \(\displaystyle\int \dfrac{1}{x(300-x)}\,dx = \int\dfrac{1}{1495}\,dt\) |
| Correct integration of \(t\) term | B1 | E.g. obtain \(t\) or \(\dfrac{t}{1495}\) |
| State or imply partial fractions of the form \(\dfrac{A}{x} + \dfrac{B}{300-x}\) | B1 | |
| Correct method to find \(A\) or \(B\) | M1 | \(A = \dfrac{1}{300}\) and \(B = \dfrac{1}{300}\); may see \(A = B = \dfrac{1495}{300} = \dfrac{299}{60}\) |
| Obtain terms \(\dfrac{1495}{300}\ln x - \dfrac{1495}{300}\ln(300-x)\) | A1 | OE. May see \(\dfrac{1}{300}\ln x - \dfrac{1}{300}\ln(300-x)\) |
| Use \(t=0\), \(x=1\) to evaluate constant or as limits in solution containing \(\ln x\), \(\ln(300-x)\) and \(t\) | M1 | |
| Obtain correct answer in any form | A1 | E.g. \(\dfrac{1495}{300}[\ln x - \ln(300-x)] = t - \dfrac{1495}{300}\ln 299\) |
| Use law of logarithms twice to obtain an expression for \(t\) | M1 | |
| Obtain \(t = \dfrac{299}{60}\ln\dfrac{299x}{300-x}\) or equivalent single logarithm | A1 | |
| 9 |
## Question 11:
**Part 11(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply equation of the form $\dfrac{dx}{dt} = kx(300-x)$ and use $\dfrac{dx}{dt} = 0.2$ and $x=1$ | M1 | M0 for verification |
| Obtain $k = \dfrac{1}{1495}$ and rearrange to given answer | A1 | $1495\dfrac{dx}{dt} = x(300-x)$ |
| | **2** | |
**Part 11(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly | B1 | $\displaystyle\int \dfrac{1}{x(300-x)}\,dx = \int\dfrac{1}{1495}\,dt$ |
| Correct integration of $t$ term | B1 | E.g. obtain $t$ or $\dfrac{t}{1495}$ |
| State or imply partial fractions of the form $\dfrac{A}{x} + \dfrac{B}{300-x}$ | B1 | |
| Correct method to find $A$ or $B$ | M1 | $A = \dfrac{1}{300}$ and $B = \dfrac{1}{300}$; may see $A = B = \dfrac{1495}{300} = \dfrac{299}{60}$ |
| Obtain terms $\dfrac{1495}{300}\ln x - \dfrac{1495}{300}\ln(300-x)$ | A1 | OE. May see $\dfrac{1}{300}\ln x - \dfrac{1}{300}\ln(300-x)$ |
| Use $t=0$, $x=1$ to evaluate constant or as limits in solution containing $\ln x$, $\ln(300-x)$ and $t$ | M1 | |
| Obtain correct answer in any form | A1 | E.g. $\dfrac{1495}{300}[\ln x - \ln(300-x)] = t - \dfrac{1495}{300}\ln 299$ |
| Use law of logarithms twice to obtain an expression for $t$ | M1 | |
| Obtain $t = \dfrac{299}{60}\ln\dfrac{299x}{300-x}$ or equivalent single logarithm | A1 | |
| | **9** | |11 In a field there are 300 plants of a certain species, all of which can be infected by a particular disease. At time $t$ after the first plant is infected there are $x$ infected plants. The rate of change of $x$ is proportional to the product of the number of plants infected and the number of plants that are not yet infected. The variables $x$ and $t$ are treated as continuous, and it is given that $\frac { \mathrm { dx } } { \mathrm { dt } } = 0.2$ and $x = 1$ when $t = 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ and $t$ satisfy the differential equation
$$1495 \frac { \mathrm { dx } } { \mathrm { dt } } = x ( 300 - x )$$
\item Using partial fractions, solve the differential equation and obtain an expression for $t$ in terms of a single logarithm involving $x$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q11 [11]}}