AQA FP3 2010 January — Question 1 8 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2010
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeApply improved Euler method for differential equations
DifficultyStandard +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—no derivation, problem-solving insight, or conceptual understanding beyond basic substitution is required. While it's an FP3 topic, the execution is purely mechanical.
Spec1.09d Newton-Raphson method

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x \ln ( 2 x + y )$$ and $$y ( 3 ) = 2$$
  1. 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 \(y ( 3.1 )\), giving your answer to four decimal places.
  2. 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 \(h = 0.1\), to obtain an approximation to \(y ( 3.1 )\), giving your answer to four decimal places.

(a)
\(y_1 = 2 + 0.1 \times [3\ln(2 \times 3 + 2)] = 2 + 0.3\ln8 = 2.6238\) (to 4dp)
AnswerMarks Guidance
\(y(3.1) = 2.6238\)M1A1, A1 Condone greater accuracy; 3 marks total
(b)
\(k_1 = 0.1 \times 3\ln 8 = 0.6238(32\ldots)\)
\(k_2 = 0.1 \times f(3.1, 2.6238(32\ldots))\)
\(\ldots = 0.1 \times 3.1 \times \ln[6.2 + \text{answer(a)}]\)
\([= 0.6750(1\ldots)]\)
\(y(3.1) = 2 + \frac{1}{2}[0.6238(3\ldots) + 0.6750(1\ldots)]\)
AnswerMarks Guidance
\(= 2.6494(2\ldots) = 2.6494\) to 4dpm1, A1 PI: ft from (a), 4dp or better; PI: ft on \(0.1 \times 3.1 \times \ln[6.2 + \text{answer(a)}]\); CAO Must be 2.6494; 5 marks total
Total: 8 marks
**(a)**
$y_1 = 2 + 0.1 \times [3\ln(2 \times 3 + 2)] = 2 + 0.3\ln8 = 2.6238$ (to 4dp)
$y(3.1) = 2.6238$ | M1A1, A1 | Condone greater accuracy; 3 marks total

**(b)**
$k_1 = 0.1 \times 3\ln 8 = 0.6238(32\ldots)$
$k_2 = 0.1 \times f(3.1, 2.6238(32\ldots))$
$\ldots = 0.1 \times 3.1 \times \ln[6.2 + \text{answer(a)}]$
$[= 0.6750(1\ldots)]$
$y(3.1) = 2 + \frac{1}{2}[0.6238(3\ldots) + 0.6750(1\ldots)]$
$= 2.6494(2\ldots) = 2.6494$ to 4dp | m1, A1 | PI: ft from (a), 4dp or better; PI: ft on $0.1 \times 3.1 \times \ln[6.2 + \text{answer(a)}]$; CAO Must be 2.6494; 5 marks total

**Total: 8 marks**

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1 The function $y ( x )$ satisfies the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$

where

$$\mathrm { f } ( x , y ) = x \ln ( 2 x + y )$$

and

$$y ( 3 ) = 2$$
\begin{enumerate}[label=(\alph*)]
\item Use the Euler formula

$$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$

with $h = 0.1$, to obtain an approximation to $y ( 3.1 )$, 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 $h = 0.1$, to obtain an approximation to $y ( 3.1 )$, giving your answer to four decimal places.
\end{enumerate}

\hfill \mbox{\textit{AQA FP3 2010 Q1 [8]}}