AQA FP3 2007 June — Question 2 9 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2007
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeIterative/numerical methods
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 conceptual insight or problem-solving is required. While it's a Further Maths topic, the mechanical nature and provision of all formulas makes it easier than an average A-level question.
Spec1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

2 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { x ^ { 2 } + y ^ { 2 } + 3 }$$ and $$y ( 1 ) = 2$$
  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 \(y ( 1.1 )\), 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 \(y ( 1.1 )\), giving your answer to four decimal places.

Question 2:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y_1 = 2 + 0.1 \times \sqrt{1^2 + 2^2 + 3}\)M1
\(y(1.1) = 2 + 0.1 \times \sqrt{8}\)A1
\(y(1.1) = 2.28284\ldots = 2.2828\) to 4dpA1 3 marks
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(k_1 = 0.1 \times \sqrt{8} = 0.2828\)M1, A1ft PI
\(k_2 = 0.1 \times f(1.1,\ 2.2828\ldots)\)M1
\(= 0.1 \times \sqrt{9.42137\ldots} = 0.3069(425\ldots)\)A1 PI
\(y(1.1) = y(1) + \frac{1}{2}[0.28284\ldots + 0.30694\ldots]\)m1
\(2.29489\ldots = 2.2949\) to 4dpA1 6 marks
## Question 2:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y_1 = 2 + 0.1 \times \sqrt{1^2 + 2^2 + 3}$ | M1 | |
| $y(1.1) = 2 + 0.1 \times \sqrt{8}$ | A1 | |
| $y(1.1) = 2.28284\ldots = 2.2828$ to 4dp | A1 | **3 marks** |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $k_1 = 0.1 \times \sqrt{8} = 0.2828$ | M1, A1ft | PI |
| $k_2 = 0.1 \times f(1.1,\ 2.2828\ldots)$ | M1 | |
| $= 0.1 \times \sqrt{9.42137\ldots} = 0.3069(425\ldots)$ | A1 | PI |
| $y(1.1) = y(1) + \frac{1}{2}[0.28284\ldots + 0.30694\ldots]$ | m1 | |
| $2.29489\ldots = 2.2949$ to 4dp | A1 | **6 marks** |

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

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

where

$$\mathrm { f } ( x , y ) = \sqrt { x ^ { 2 } + y ^ { 2 } + 3 }$$

and

$$y ( 1 ) = 2$$
\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 $y ( 1.1 )$, 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 $y ( 1.1 )$, giving your answer to four decimal places.
\end{enumerate}

\hfill \mbox{\textit{AQA FP3 2007 Q2 [9]}}