| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable with partial fractions |
| Difficulty | Standard +0.3 This is a straightforward separable differential equation requiring partial fractions decomposition of 1/((20-x)(40-x)), followed by integration and applying initial conditions. While it involves multiple steps (separation, partial fractions, integration, solving for x), each technique is standard A-level material with no novel insight required. The 9 marks reflect routine algebraic manipulation rather than conceptual difficulty, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly and integrate one side | B1 | |
| Obtain term \(0.2t\), or equivalent | B1 | |
| Carry out a relevant method to obtain \(A\) and \(B\) such that \(\frac{1}{(20-x)(40-x)} \equiv \frac{A}{20-x} + \frac{B}{40-x}\) | \*M1 | OE |
| Obtain \(A = \frac{1}{20}\) and \(B = -\frac{1}{20}\) | A1 | |
| Integrate and obtain terms \(-\frac{1}{20}\ln(20-x) + \frac{1}{20}\ln(40-x)\) OE | A1FT +A1FT | The FT is on \(A\) and \(B\) |
| Use \(x = 10\), \(t = 0\) to evaluate a constant, or as limits | DM1 | |
| Obtain correct answer in any form | A1 | |
| Obtain final answer \(x = \frac{60e^{4t}-40}{3e^{4t}-1}\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State that \(x\) approaches 20 | B1 |
## Question 9(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly and integrate one side | B1 | |
| Obtain term $0.2t$, or equivalent | B1 | |
| Carry out a relevant method to obtain $A$ and $B$ such that $\frac{1}{(20-x)(40-x)} \equiv \frac{A}{20-x} + \frac{B}{40-x}$ | \*M1 | OE |
| Obtain $A = \frac{1}{20}$ and $B = -\frac{1}{20}$ | A1 | |
| Integrate and obtain terms $-\frac{1}{20}\ln(20-x) + \frac{1}{20}\ln(40-x)$ OE | A1FT +A1FT | The FT is on $A$ and $B$ |
| Use $x = 10$, $t = 0$ to evaluate a constant, or as limits | DM1 | |
| Obtain correct answer in any form | A1 | |
| Obtain final answer $x = \frac{60e^{4t}-40}{3e^{4t}-1}$ | A1 | OE |
**Total: 9 marks**
---
## Question 9(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| State that $x$ approaches 20 | B1 | |
**Total: 1 mark**
---9 The variables $x$ and $t$ satisfy the differential equation $5 \frac { \mathrm {~d} x } { \mathrm {~d} t } = ( 20 - x ) ( 40 - x )$. It is given that $x = 10$ when $t = 0$.\\
(i) Using partial fractions, solve the differential equation, obtaining an expression for $x$ in terms of $t$. [9]\\
(ii) State what happens to the value of $x$ when $t$ becomes large.\\
\hfill \mbox{\textit{CAIE P3 2019 Q9 [10]}}