FP2 June 2018 Q5

EdexcelOld spec9 marksTaylor Series

5. \[y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\frac{\mathrm{d}y}{\mathrm{d}x} - 3y^2 = 0\]

Given that at \(x = 0\), \(y = 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\)

(a) show that, at \(x = 0\), \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{3}{2}\) (6)

(b) Find a series solution for \(y\) up to and including the term in \(x^3\) (3)

Mark scheme (a)

Scheme	Marks
\(y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3y^2 = 0\)
\(y\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + \dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) M1: Use of Product Rule on \(y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\), 2 terms added with at least one term correct. A1: Fully correct derivative of \(y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)	M1,A1
\(+3x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) Correct derivative of \(3x\dfrac{\mathrm{d}y}{\mathrm{d}x}\)	B1
\(-6y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe.	B1
At \(x = 0\), \(2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3(0)(1) - 3(4) = 0 \Rightarrow \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \ldots\) and \(2\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + (1)(6) + 3(1) - 6(2)(1) = 0 \Rightarrow \dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \ldots\) Sub \(x = 0,\ y = 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) (must use these values) leading to numerical values for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\)	M1
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{3}{2}\) Given answer** cso	A1cso
	(6)

Scheme	Marks
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + \dfrac{3x}{y}\cdot\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3y = 0\)
\(\left(\dfrac{3y - 3x\dfrac{\mathrm{d}y}{\mathrm{d}x}}{y^2}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{3x}{y}\times\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) oe M1 Use of Product Rule on \(\dfrac{3x}{y}\times\dfrac{\mathrm{d}y}{\mathrm{d}x}\), 2 terms added with at least one term correct A1 Correct derivative	M1A1
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) oe	B1
\(-3\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe	B1
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + \left(\dfrac{3\times2 - 0}{2^2}\right)\times1 + 0\times\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 3\times1 \to \dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \ldots\) Sub \(x = 0,\ y = 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) (must use these values) leading to numerical value for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) (value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) not needed)	M1
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{3}{2}\) Given answer** cso	A1cso

Scheme	Marks
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{1}{y}\left(3y^2 - 3\dfrac{\mathrm{d}y}{\mathrm{d}x} x\right)\)
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = +\dfrac{1}{y}\left(6y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)\), \(-\dfrac{1}{y^2}\dfrac{\mathrm{d}y}{\mathrm{d}x}\left(3y^2 - 3x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\) B1 \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\), M1 Differentiate using product rule. 2 terms added with at least one term correct A1: \(+\dfrac{1}{y}\left(6y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)\) B1 \(-\dfrac{1}{y^2}\dfrac{\mathrm{d}y}{\mathrm{d}x}\left(3y^2 - 3x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\)	B1, M1A1, B1
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{1}{2}(6\times2\times1 - 3\times1 - 3\times0\times6)\) \(-\dfrac{1}{4}\times1(3\times4 - 3\times0\times1) \Rightarrow \dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \ldots\) Sub \(x = 0,\ y = 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) (must use these values) leading to numerical value for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) (value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) not needed)	M1
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{3}{2}\) Given answer** cso	A1cso

Mark scheme (b)

Scheme	Marks
\((y =)\ 2 + x\) Use the given values to form the first 2 terms of the series	B1
\((y =)\ 2 + x + \dfrac{6}{2!}x^2 + \dfrac{\frac{3}{2}}{3!}x^3\ (+\ldots)\) Find a numerical value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) (may be seen in (a)) and use with the given value of \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) to form the \(x^2\) and \(x^3\) terms of the series expansion	M1
\(y = 2 + x + 3x^2 + \dfrac{1}{4}x^3\ (+\ldots)\) Follow through their value of \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) used correctly. Must start \(y = \ldots\) Allow \(\mathrm{f}(x)\) only if this has been defined anywhere in the question to be equal to \(y\)	A1ft
	(3)
	(9 marks)