AQA FP3 2008 January — Question 1 6 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2008
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeApply basic Euler method for differential equations
DifficultyStandard +0.3 This is a straightforward application of two standard numerical methods (Euler's method and a mid-point variant) with explicit formulas provided. Students simply substitute given values into the formulas with h=0.1, requiring only basic arithmetic and function evaluation. No conceptual insight or problem-solving is needed beyond following the given procedures, making it slightly easier than average.
Spec1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x ^ { 2 } - y ^ { 2 }$$ and $$y ( 2 ) = 1$$
  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 \(y ( 2.1 )\).
  2. 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), to obtain an approximation to \(y ( 2.2 )\).

AnswerMarks Guidance
(a) \(y(2,1) = y(2) + 0.1[2^2 - 1] = 1+0.1 \times 3 = 1.3\)M1A1, A1 Total: 3
(b) \(y(2.2) = y(2) + 2(0.1)[f(2.1, y(2.1))]\)M1
\(\ldots = 1+2(0.1)[2.1^2 - 1.3^2]\)A1/
\(\ldots = 1+0.2 \times 2.72 = 1.544\)A1 Total: 3
**(a)** $y(2,1) = y(2) + 0.1[2^2 - 1] = 1+0.1 \times 3 = 1.3$ | M1A1, A1 | Total: 3 | Ft on cand's answer to (a)

**(b)** $y(2.2) = y(2) + 2(0.1)[f(2.1, y(2.1))]$ | M1 | | 
$\ldots = 1+2(0.1)[2.1^2 - 1.3^2]$ | A1/ | | Ft on cand's answer to (a)
$\ldots = 1+0.2 \times 2.72 = 1.544$ | A1 | Total: 3 | CAO
1 The function $y ( x )$ satisfies the differential equation

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

where

$$\mathrm { f } ( x , y ) = x ^ { 2 } - y ^ { 2 }$$

and

$$y ( 2 ) = 1$$
\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 $y ( 2.1 )$.
\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), to obtain an approximation to $y ( 2.2 )$.
\end{enumerate}

\hfill \mbox{\textit{AQA FP3 2008 Q1 [6]}}