| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Standard +0.8 This is a multi-part separable differential equation requiring separation of variables, integration involving logarithms and exponentials, application of initial conditions, and a limiting argument. While the separation is straightforward, the algebraic manipulation and the final limit analysis (part iii) require careful reasoning beyond routine textbook exercises, placing it moderately above average difficulty. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Separate variables correctly and attempt integration of one side | B1 | |
| Obtain term \(\ln x\) | B1 | |
| Obtain term of the form \(a\ln(k + e^{-t})\) | M1 | |
| Obtain term \(-\ln(k + e^{-t})\) | A1 | |
| Evaluate a constant or use limits \(x = 10\), \(t = 0\) in a solution containing terms \(a\ln(k + e^{-t})\) and \(b\ln x\) | M1✱ | |
| Obtain correct solution in any form, e.g. \(\ln x - \ln 10 = -\ln(k + e^{-t}) + \ln(k + 1)\) | A1 | [6] |
| (ii) Substitute \(x = 20\), \(t = 1\) and solve for \(k\) | M1(dep✱) | |
| Obtain the given answer | A1 | [2] |
| (iii) Using \(e^{-t} \to 0\) and the given value of \(k\), find the limiting value of \(x\) | M1 | |
| Justify the given answer | A1 | [2] |
**(i)** Separate variables correctly and attempt integration of one side | B1 |
Obtain term $\ln x$ | B1 |
Obtain term of the form $a\ln(k + e^{-t})$ | M1 |
Obtain term $-\ln(k + e^{-t})$ | A1 |
Evaluate a constant or use limits $x = 10$, $t = 0$ in a solution containing terms $a\ln(k + e^{-t})$ and $b\ln x$ | M1✱ |
Obtain correct solution in any form, e.g. $\ln x - \ln 10 = -\ln(k + e^{-t}) + \ln(k + 1)$ | A1 | [6]
**(ii)** Substitute $x = 20$, $t = 1$ and solve for $k$ | M1(dep✱) |
Obtain the given answer | A1 | [2]
**(iii)** Using $e^{-t} \to 0$ and the given value of $k$, find the limiting value of $x$ | M1 |
Justify the given answer | A1 | [2]9 The number of organisms in a population at time $t$ is denoted by $x$. Treating $x$ as a continuous variable, the differential equation satisfied by $x$ and $t$ is
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { x \mathrm { e } ^ { - t } } { k + \mathrm { e } ^ { - t } }$$
where $k$ is a positive constant.\\
(i) Given that $x = 10$ when $t = 0$, solve the differential equation, obtaining a relation between $x , k$ and $t$.\\
(ii) Given also that $x = 20$ when $t = 1$, show that $k = 1 - \frac { 2 } { \mathrm { e } }$.\\
(iii) Show that the number of organisms never reaches 48, however large $t$ becomes.
\hfill \mbox{\textit{CAIE P3 2015 Q9 [10]}}