| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Iterative/numerical methods |
| Difficulty | Standard +0.3 This is a standard Further Maths question combining numerical methods (Euler's method) with integrating factor technique. Part (a) involves straightforward substitution into given formulas. Part (b) requires recognizing the standard form, verifying the integrating factor (routine check), and solving by integration—all standard FP3 techniques with no novel insight required. Slightly above average difficulty due to being Further Maths content, but entirely procedural. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y(1.1) = y(1) + 0.1[1\cdot\ln 1 + 1/1]\) | M1A1 | |
| \(= 1 + 0.1 = 1.1\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y(1.2) = y(1) + 2(0.1)[f(1.1,\, y(1.1))]\) | M1A1 | |
| \(\ldots = 1 + 2(0.1)[1.1\ln 1.1 + (1.1)/1.1]\) | A1\(\checkmark\) | On answer to (a)(i) |
| \(\ldots = 1 + 0.2 \times 1.104841198\ldots\) | ||
| \(\ldots = 1.22096824\ldots = 1.221\) to 3dp | A1 | 4 marks; CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| IF is \(e^{\int -\frac{1}{x}\,dx}\) | M1 | Condone \(e^{\int\frac{1}{x}\,dx}\) for M mark |
| \(= e^{-\ln x}\) | A1 | |
| \(= e^{\ln x^{-1}} = x^{-1} = \frac{1}{x}\) | A1 | 3 marks; AG (be convinced); Solutions using printed answer must be convincing before any marks are awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dx}\left(\frac{y}{x}\right) = \ln x\) | M1A1 | |
| \(\frac{y}{x} = \int \ln x\,dx = x\ln x - \int x\left(\frac{1}{x}\right)dx\) | M1 | Integration by parts for \(x^k\ln x\) |
| \(\frac{y}{x} = x\ln x - x + c\) | A1 | Condone missing \(c\) |
| \(y(1) = 1 \Rightarrow 1 = \ln 1 - 1 + c\) | m1 | Dependent on at least one of the two previous M marks |
| \(\Rightarrow c = 2 \Rightarrow y = x^2\ln x - x^2 + 2x\) | A1 | 6 marks; OE e.g. \(\frac{y}{x} = x\ln x - x + 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y(1.2) = 1.222543\ldots = 1.223\) to 3dp | B1 | 1 mark |
# Question 5:
## Part 5(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y(1.1) = y(1) + 0.1[1\cdot\ln 1 + 1/1]$ | M1A1 | |
| $= 1 + 0.1 = 1.1$ | A1 | 3 marks |
## Part 5(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y(1.2) = y(1) + 2(0.1)[f(1.1,\, y(1.1))]$ | M1A1 | |
| $\ldots = 1 + 2(0.1)[1.1\ln 1.1 + (1.1)/1.1]$ | A1$\checkmark$ | On answer to (a)(i) |
| $\ldots = 1 + 0.2 \times 1.104841198\ldots$ | | |
| $\ldots = 1.22096824\ldots = 1.221$ to 3dp | A1 | 4 marks; CAO |
## Part 5(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| IF is $e^{\int -\frac{1}{x}\,dx}$ | M1 | Condone $e^{\int\frac{1}{x}\,dx}$ for M mark |
| $= e^{-\ln x}$ | A1 | |
| $= e^{\ln x^{-1}} = x^{-1} = \frac{1}{x}$ | A1 | 3 marks; AG (be convinced); Solutions using printed answer must be convincing before any marks are awarded |
## Part 5(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}\left(\frac{y}{x}\right) = \ln x$ | M1A1 | |
| $\frac{y}{x} = \int \ln x\,dx = x\ln x - \int x\left(\frac{1}{x}\right)dx$ | M1 | Integration by parts for $x^k\ln x$ |
| $\frac{y}{x} = x\ln x - x + c$ | A1 | Condone missing $c$ |
| $y(1) = 1 \Rightarrow 1 = \ln 1 - 1 + c$ | m1 | Dependent on at least one of the two previous M marks |
| $\Rightarrow c = 2 \Rightarrow y = x^2\ln x - x^2 + 2x$ | A1 | 6 marks; OE e.g. $\frac{y}{x} = x\ln x - x + 2$ |
## Part 5(b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y(1.2) = 1.222543\ldots = 1.223$ to 3dp | B1 | 1 mark |5
\begin{enumerate}[label=(\alph*)]
\item The function $y ( x )$ satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$
where
$$\mathrm { f } ( x , y ) = x \ln x + \frac { y } { x }$$
and
$$y ( 1 ) = 1$$
\begin{enumerate}[label=(\roman*)]
\item Use the Euler formula
$$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$
with $h = 0.1$, to obtain an approximation to $y ( 1.1 )$.
\item Use the formula
$$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$
with your answer to part (a)(i) to obtain an approximation to $y ( 1.2 )$, giving your answer to three decimal places.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that $\frac { 1 } { x }$ is an integrating factor for the first-order differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { 1 } { x } y = x \ln x$$
\item Solve this differential equation, given that $y = 1$ when $x = 1$.
\item Calculate the value of $y$ when $x = 1.2$, giving your answer to three decimal places.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2006 Q5 [17]}}