Edexcel FP2 2012 June — Question 5 10 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeExplicit differential equation series solution
DifficultyChallenging +1.2 This is a standard FP2 Taylor series question requiring differentiation of the given DE to find higher derivatives, then substitution of the initial condition to build the series term-by-term. Part (a) is routine implicit differentiation, and part (b) follows a well-practiced algorithm. While it requires careful algebraic manipulation and is harder than typical C3/C4 content due to being Further Maths, it's a textbook exercise without novel problem-solving demands.
Spec1.07q Product and quotient rules: differentiation4.08a Maclaurin series: find series for function

5. $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 x + y ^ { 2 }$$
  1. Show that $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 1 - 2 y ) \frac { \mathrm { d } y } { \mathrm {~d} x } = 3$$ Given that \(y = 1\) at \(x = 1\),
  2. find a series solution for \(y\) in ascending powers of ( \(x - 1\) ), up to and including the term in \(( x - 1 ) ^ { 3 }\).

5.

$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 x + y ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that

$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 1 - 2 y ) \frac { \mathrm { d } y } { \mathrm {~d} x } = 3$$

Given that $y = 1$ at $x = 1$,
\item find a series solution for $y$ in ascending powers of ( $x - 1$ ), up to and including the term in $( x - 1 ) ^ { 3 }$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2 2012 Q5 [10]}}