Edexcel FP2 2006 June — Question 7 11 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2006
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeExplicit differential equation series solution
DifficultyChallenging +1.2 This is a standard Further Maths FP2 technique: using Taylor series to solve a differential equation with initial conditions. Students differentiate the DE repeatedly, substitute initial conditions, and build the series term-by-term. While it requires careful algebraic manipulation and understanding of the small-angle approximation (sin x ≈ x), it's a well-practiced exam technique with a clear algorithmic procedure. The question is moderately harder than average A-level due to being Further Maths content, but it's routine within FP2.
Spec4.08a Maclaurin series: find series for function

7. $$\frac { \mathrm { d } ^ { 2 x } } { \mathrm {~d} t ^ { 2 } } + 3 \sin x = 0 . \quad \text { At } t = 0 , \quad x = 0 \quad \text { and } \quad \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4$$ (b) Find a series solution for \(x\), in ascending powers of \(t\), up to and including the term in \(t ^ { 3 }\).
(c) Use your answer to (b) to obtain an estimate of \(x\) at \(t = 0.3\).
(2)(Total 11 marks)

AnswerMarks Guidance
(a) \(\left(\frac{dx}{dt}\right)_0 = 0.4 \approx \frac{x_{0,1}-0}{0.1} \Rightarrow x_{0,1} \approx 0.04\)B1
\(\left(\frac{d^2x}{dt^2}\right)_{0,1} = 3\sin x_{0,1} \approx \frac{x_{0,2}-2x_{0,1}+0}{0.01}\)M1
Must have their \(x_{0,1}\)
\(x_{0,2} \approx 0.0788\) awrtA1
\(\left(\frac{d^2x}{dt^2}\right)_{0,2} = 3\sin x_{0,2} \approx \frac{x_{0,3}-2x_{0,2}+x_{0,1}}{0.01}\)M1
Must have their \(x_{0,1}, x_{0,2}\)
\(x_{0,3} \approx 0.115\) awrtA1 5 marks
(b) \(f''(t) = -3\sin x\), \(f''(0) = 0\)
\(f'''(t) = -3\cos x \frac{dx}{dt}\), \(f'''(0) = -3 \times 0.4 = -1.2\)M1 A1
\(f(t) = f(0) + f'(0) \frac{t^2}{2}f''(0) + \frac{t^3}{3!}f'''(0) + \ldots\)
\(= 0.4t - 0.2t^3\)M1 A1 4 marks
(c) Substituting \(t = 0.3\) into their answer to (b) and evaluating \(f(0.3) = 0.1146\) caoM1 A1 2 marks [11]
**(a)** $\left(\frac{dx}{dt}\right)_0 = 0.4 \approx \frac{x_{0,1}-0}{0.1} \Rightarrow x_{0,1} \approx 0.04$ | B1 |

$\left(\frac{d^2x}{dt^2}\right)_{0,1} = 3\sin x_{0,1} \approx \frac{x_{0,2}-2x_{0,1}+0}{0.01}$ | M1 |

Must have their $x_{0,1}$ | |

$x_{0,2} \approx 0.0788$ awrt | A1 |

$\left(\frac{d^2x}{dt^2}\right)_{0,2} = 3\sin x_{0,2} \approx \frac{x_{0,3}-2x_{0,2}+x_{0,1}}{0.01}$ | M1 |

Must have their $x_{0,1}, x_{0,2}$ | |

$x_{0,3} \approx 0.115$ awrt | A1 | 5 marks

**(b)** $f''(t) = -3\sin x$, $f''(0) = 0$ | |

$f'''(t) = -3\cos x \frac{dx}{dt}$, $f'''(0) = -3 \times 0.4 = -1.2$ | M1 A1 |

$f(t) = f(0) + f'(0) \frac{t^2}{2}f''(0) + \frac{t^3}{3!}f'''(0) + \ldots$ | |

$= 0.4t - 0.2t^3$ | M1 A1 | 4 marks

**(c)** Substituting $t = 0.3$ into their answer to (b) and evaluating $f(0.3) = 0.1146$ cao | M1 A1 | 2 marks [11]

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7.

$$\frac { \mathrm { d } ^ { 2 x } } { \mathrm {~d} t ^ { 2 } } + 3 \sin x = 0 . \quad \text { At } t = 0 , \quad x = 0 \quad \text { and } \quad \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4$$

(b) Find a series solution for $x$, in ascending powers of $t$, up to and including the term in $t ^ { 3 }$.\\
(c) Use your answer to (b) to obtain an estimate of $x$ at $t = 0.3$.\\
(2)(Total 11 marks)\\

\hfill \mbox{\textit{Edexcel FP2 2006 Q7 [11]}}