| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| 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 straightforward application of two standard numerical methods (Euler and improved Euler) with explicit formulas provided. Students only need to substitute given values into the formulas and perform arithmetic calculations with a calculator. No conceptual insight or problem-solving is required beyond following the given procedures, making it slightly easier than average for Further Maths. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y(1.05) = 0.6 + 0.05 \times [\ln(1 + 1 + 0.6)]\) | M1A1 | |
| \(= 0.6477(7557...) = 0.6478\) to 4dp | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(k_1 = 0.05 \times \ln(1 + 1 + 0.6) = 0.0477(75...)\) | M1 | PI |
| A1F | ft candidate's evaluation in (a) | |
| \(k_2 = 0.05 \times f(1.05,\ 0.6477...)\) | M1 | |
| \(... = 0.05 \times \ln(1 + 1.05^2 + 0.6477...)\) | ||
| \(... = 0.0505(85...)\) | A1F | PI |
| \(y(1.05) = y(1) + \frac{1}{2}[k_1 + k_2]\) | m1 | Dep on previous two Ms and numerical values for \(k\)'s |
| \(= 0.6 + 0.5 \times 0.09836...\) | ||
| \(= 0.6492\) to 4dp | A1F | Total: 6 |
## Question 1:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $y(1.05) = 0.6 + 0.05 \times [\ln(1 + 1 + 0.6)]$ | M1A1 | |
| $= 0.6477(7557...) = 0.6478$ to 4dp | A1 | Total: 3 | Condone >4 dp |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $k_1 = 0.05 \times \ln(1 + 1 + 0.6) = 0.0477(75...)$ | M1 | PI |
| | A1F | ft candidate's evaluation in (a) |
| $k_2 = 0.05 \times f(1.05,\ 0.6477...)$ | M1 | |
| $... = 0.05 \times \ln(1 + 1.05^2 + 0.6477...)$ | | |
| $... = 0.0505(85...)$ | A1F | PI |
| $y(1.05) = y(1) + \frac{1}{2}[k_1 + k_2]$ | m1 | Dep on previous two Ms and numerical values for $k$'s |
| $= 0.6 + 0.5 \times 0.09836...$ | | |
| $= 0.6492$ to 4dp | A1F | Total: 6 | Must be 4 dp... ft one slip |
---1 The function $y ( x )$ satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$
where
$$\mathrm { f } ( x , y ) = \ln \left( 1 + x ^ { 2 } + y \right)$$
and
$$y ( 1 ) = 0.6$$
\begin{enumerate}[label=(\alph*)]
\item Use the Euler formula
$$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$
with $h = 0.05$, to obtain an approximation to $y ( 1.05 )$, giving your answer to four decimal places.
\item Use the improved Euler formula
$$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$
where $k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)$ and $k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)$ and $h = 0.05$, to obtain an approximation to $y ( 1.05 )$, giving your answer to four decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2007 Q1 [9]}}