Edexcel FP2 — Question 2 7 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Marks7
PaperDownload PDF ↗
TopicTaylor series
TypeDifferential equation given
DifficultyStandard +0.8 This is a Further Maths FP2 question requiring repeated differentiation of an implicit differential equation using product and chain rules, then constructing a Taylor series. Part (a) demands careful algebraic manipulation through multiple derivatives, while part (b) requires systematic evaluation at x=0 and series construction. The complexity of the expressions and multi-step nature elevates this above standard A-level, though it follows a predictable Taylor series methodology for FP2.
Spec4.08a Maclaurin series: find series for function

$$\frac{d^2 y}{dx^2} = e^x \left(2x \frac{dy}{dx} + y^2 + 1\right).$$
  1. Show that $$\frac{d^4 y}{dx^4} = e^x \left[2x \frac{d^3 y}{dx^3} + 4 \frac{d^2 y}{dx^2} + 6y \frac{dy}{dx} + y^2 + 1\right],$$ where \(k\) is a constant to be found. [3]
Given that, at \(x = 0\), \(y = 1\) and \(\frac{dy}{dx} = 2\),
  1. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x^4\). [4]

$$\frac{d^2 y}{dx^2} = e^x \left(2x \frac{dy}{dx} + y^2 + 1\right).$$

\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{d^4 y}{dx^4} = e^x \left[2x \frac{d^3 y}{dx^3} + 4 \frac{d^2 y}{dx^2} + 6y \frac{dy}{dx} + y^2 + 1\right],$$
where $k$ is a constant to be found. [3]
\end{enumerate}

Given that, at $x = 0$, $y = 1$ and $\frac{dy}{dx} = 2$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find a series solution for $y$ in ascending powers of $x$, up to and including the term in $x^4$. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2  Q2 [7]}}