FP2 June 2005 Q11

EdexcelOld spec8 marksTaylor Series

11. The variable \(y\) satisfies the differential equation \[4(1 + x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4x\frac{\mathrm{d}y}{\mathrm{d}x} = y.\] At \(x = 0\), \(y = 1\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\).

(a) Find the value of \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(x = 0\). (1)

(d) Express \(y\) as a series, in ascending powers of \(x\), up to and including the term in \(x^3\). (2)

(e) Find the value that the series gives for \(y\) at \(x = 0.1\), giving your answer to 5 decimal places. (1)

Scheme	Marks
\(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = \dfrac{1}{4}\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
Diff: \(4(1 + x^2)\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + 8x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}x}\)	M1A1
Substituting appropriate vales \(\Rightarrow 4\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = -\dfrac{3}{2} \Rightarrow \left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = -\dfrac{3}{8}\)	M1A1
	(4)

Scheme	Marks
\(y = y_0 + y_0'x + \dfrac{y_0''}{2!}x^2 + \dfrac{y_0'''}{3!}x^3 + \ldots = 1 + \dfrac{1}{2}x + \dfrac{1}{8}x^2 - \dfrac{1}{16}x^3 + \ldots\)	M1A1ft
	(2)