| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| 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 +0.3 Part (i) requires translating a word problem into a differential equation using proportionality and initial conditions to find the constant—straightforward application of 'rate proportional to product' with simple algebra. Parts (ii-iii) involve standard separable differential equation techniques (partial fractions, integration, applying boundary conditions) that are routine for P3 level, making this slightly easier than average overall. |
| Spec | 1.08l Interpret differential equation solutions: in context4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| State or imply \(\frac{dx}{dt} = k(10-x)(20-x)\) and show \(k = 0.01\) | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Separate variables correctly and attempt integration of at least one side | M1 | |
| Carry out an attempt to find \(A\) and \(B\) such that \(\frac{1}{(10-x)(20-x)} \equiv \frac{A}{10-x} + \frac{B}{20-x}\), or equivalent | M1 | |
| Obtain \(A = \frac{1}{10}\) and \(B = -\frac{1}{10}\), or equivalent | A1 | |
| Integrate and obtain \(-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x)\), or equivalent | A1√ | |
| Integrate and obtain term 0.01t, or equivalent | A1 | |
| Evaluate a constant, or use limits \(t = 0, x = 0\), in a solution containing terms of the form \(a\ln(10-x)\), \(b\ln(20-x)\) and \(ct\) | M1 | |
| Obtain answer in any form, e.g. \(-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x) = 0.01t + \frac{1}{10}\ln 2\) | A1√ | |
| Use laws of logarithms to correctly remove logarithms | M1 | |
| Rearrange and obtain \(x = 20(\exp(0.1t) - 1)/(2\exp(0.1t) - 1)\), or equivalent | A1 | [9] |
| Answer | Marks | Guidance |
|---|---|---|
| State that \(x\) approaches 10 | B1 | [1] |
**(i)**
State or imply $\frac{dx}{dt} = k(10-x)(20-x)$ and show $k = 0.01$ | B1 | [1]
**(ii)**
Separate variables correctly and attempt integration of at least one side | M1 |
Carry out an attempt to find $A$ and $B$ such that $\frac{1}{(10-x)(20-x)} \equiv \frac{A}{10-x} + \frac{B}{20-x}$, or equivalent | M1 |
Obtain $A = \frac{1}{10}$ and $B = -\frac{1}{10}$, or equivalent | A1 |
Integrate and obtain $-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x)$, or equivalent | A1√ |
Integrate and obtain term 0.01t, or equivalent | A1 |
Evaluate a constant, or use limits $t = 0, x = 0$, in a solution containing terms of the form $a\ln(10-x)$, $b\ln(20-x)$ and $ct$ | M1 |
Obtain answer in any form, e.g. $-\frac{1}{10}\ln(10-x) + \frac{1}{10}\ln(20-x) = 0.01t + \frac{1}{10}\ln 2$ | A1√ |
Use laws of logarithms to correctly remove logarithms | M1 |
Rearrange and obtain $x = 20(\exp(0.1t) - 1)/(2\exp(0.1t) - 1)$, or equivalent | A1 | [9]
**(iii)**
State that $x$ approaches 10 | B1 | [1]9 In a chemical reaction, a compound $X$ is formed from two compounds $Y$ and $Z$. The masses in grams of $X , Y$ and $Z$ present at time $t$ seconds after the start of the reaction are $x , 10 - x$ and $20 - x$ respectively. At any time the rate of formation of $X$ is proportional to the product of the masses of $Y$ and $Z$ present at the time. When $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 2$.\\
(i) Show that $x$ and $t$ satisfy the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.01 ( 10 - x ) ( 20 - x )$$
(ii) Solve this differential equation and obtain an expression for $x$ in terms of $t$.\\
(iii) State what happens to the value of $x$ when $t$ becomes large.
\hfill \mbox{\textit{CAIE P3 2011 Q9 [11]}}