A2 June 2025 Q2

EdexcelCurrent spec8 marksTaylor Series

\[\left[\begin{gathered}\textit{The Taylor series expansion of}\;\; \mathrm{f}(x)\;\; \textit{about}\;\; x = a\;\; \textit{is given by}\\ \mathrm{f}(x) = \mathrm{f}(a) + (x - a)\mathrm{f}^{\prime}(a) + \frac{(x - a)^2}{2!}\mathrm{f}^{\prime\prime}(a) + \ldots + \frac{(x - a)^r}{r!}\mathrm{f}^{(r)}(a) + \ldots\end{gathered}\right]\]

Given that

\[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 3\frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 4 \qquad \text{(I)}\]

(a) show that\[\frac{\mathrm{d}^4 y}{\mathrm{d}x^4} = a\frac{\mathrm{d}y}{\mathrm{d}x} + bx\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + c\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}\]where \(a\), \(b\) and \(c\) are integers to be determined. (4)

Hence, given that \(y = 1\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) when \(x = 2\)

(b) determine a Taylor series solution, in ascending powers of \((x - 2)\), up to and including the term in \((x - 2)^4\), of the differential equation (I), giving each coefficient in simplest form. (4)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + 3\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + Ax\dfrac{\mathrm{d}y}{\mathrm{d}x} + By = 0 \quad A, B \neq 0\) \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -3\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + Ax\dfrac{\mathrm{d}y}{\mathrm{d}x} + By \quad A, B \neq 0\)	M1	2.1
\(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + 3\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = 0\) or \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -3\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\)	A1	1.1b
\(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} + 3\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 2x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = \ldots\) Or \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = -3\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + 2x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = -3\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + 2x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + 4\dfrac{\mathrm{d}y}{\mathrm{d}x}\)	M1	1.1b
\(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = 4\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2x\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - 3\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) cso	A1	2.2a
	(4)

Notes

M1: Attempts to differentiate including an attempt at the product rule to achieve the correct form in any order

A1: Correct differentiation

M1: Continues to differentiate using the product rule again to reach the 4^th derivative to achieve the correct form \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = A\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} + Bx\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + C\dfrac{\mathrm{d}y}{\mathrm{d}x} + D\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in any order

A1: Completes the process to obtain the correct expression (\(a = 4\), \(b = 2\), \(c = -3\)) with no errors seen and correct notation throughout and \(= 0\) where applicable cso

Mark scheme (b)

Scheme	Marks	AO
\(x = 2, y = 1, \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 \Rightarrow \left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)_2 = 4 + 4 - 3 = 5\)	B1	2.2a
\(\left(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\right)_2 = 2(2)(1) + 2(1) - 3(5) = -9\) \(\left(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\right)_2 = 4(1) + 2(2)(5) - 3(-9) = 51\)	M1	1.1b
\(y = y(2) + (x - 2)\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_2 + \dfrac{(x - 2)^2}{2!}\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)_2 + \dfrac{(x - 2)^3}{3!}\left(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\right)_2 + \dfrac{(x - 2)^4}{4!}\left(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\right)_2 + \ldots\)	M1	2.5
\(y = 1 + (x - 2) + \dfrac{5}{2}(x - 2)^2 - \dfrac{3}{2}(x - 2)^3 + \dfrac{17}{8}(x - 2)^4 + \ldots\) Or \(y = x - 1 + \dfrac{5}{2}(x - 2)^2 - \dfrac{3}{2}(x - 2)^3 + \dfrac{17}{8}(x - 2)^4 + \ldots\)	A1	1.1b
	(4)
	(8 marks)

Notes

B1: Deduces the correct value for \(y^{\prime\prime}(2)\)

M1: Finds the values of the derivatives at \(x = 2\). There must be evidence of the correct values of \(x = 2, y = 1, \dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) being used at least once

M1: Uses the correct Taylor series expansion with their derivatives. Must be using \(y = 1, \dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) in their series expansion. If no method is shown they must have all terms correct for their derivatives

A1: Correct expansion must include \(y =\) or \(\mathrm{f}(x) =\) somewhere