Edexcel FP2 2008 June — Question 9

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeDifferential equation given
DifficultyChallenging +1.2 This is a structured Taylor series question with clear guidance through each step. Part (a) is routine differentiation using the product rule. Parts (b) and (c) involve standard techniques: substituting x=0 to find successive derivatives, then constructing the series and evaluating it. While it requires careful algebraic manipulation and understanding of Taylor series methodology, the question provides scaffolding and follows a predictable pattern typical of FP2 Taylor series problems. It's moderately harder than average A-level due to being Further Maths content, but straightforward within that context.
Spec4.08a Maclaurin series: find series for function

$$(x^2 + 1)\frac{d^2y}{dx^2} = 2y^2 + (1 - 2x)\frac{dy}{dx}$$ (I)
  1. By differentiating equation (I) with respect to \(x\), show that $$(x^2 + 1)\frac{d^3y}{dx^3} = (1 - 4x)\frac{d^2y}{dx^2} + (4y - 2)\frac{dy}{dx}.$$ (3) Given that \(y = 1\) and \(\frac{dy}{dx} = 1\) at \(x = 0\),
  2. find the series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x_3\).(4)
  3. Use your series to estimate the value of \(y\) at \(x = -0.5\), giving your answer to two decimal places.(1)

$$(x^2 + 1)\frac{d^2y}{dx^2} = 2y^2 + (1 - 2x)\frac{dy}{dx}$$ (I)

\begin{enumerate}[label=(\alph*)]
\item By differentiating equation (I) with respect to $x$, show that
$$(x^2 + 1)\frac{d^3y}{dx^3} = (1 - 4x)\frac{d^2y}{dx^2} + (4y - 2)\frac{dy}{dx}.$$ (3)

Given that $y = 1$ and $\frac{dy}{dx} = 1$ at $x = 0$,

\item find the series solution for $y$, in ascending powers of $x$, up to and including the term in $x_3$.(4)

\item Use your series to estimate the value of $y$ at $x = -0.5$, giving your answer to two decimal places.(1)
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2 2008 Q9}}