FP2 June 2013 Q7

EdexcelOld spec13 marksSecond Order Differentials

(a) Show that the transformation \(y = xv\) transforms the equation \[4x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 8x\frac{\mathrm{d}y}{\mathrm{d}x} + (8 + 4x^2)y = x^4 \qquad \text{(I)}\] into the equation \[4\frac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4v = x \qquad \text{(II)}\] (6)

(b) Solve the differential equation (II) to find \(v\) as a function of \(x\). (6)

Mark scheme (a)

Scheme	Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\)	M1
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{\mathrm{d}v}{\mathrm{d}x} + \dfrac{\mathrm{d}v}{\mathrm{d}x} + x\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2}\)	M1A1
\(4x^2\left(2\dfrac{\mathrm{d}v}{\mathrm{d}x} + x\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2}\right) - 8x\left(v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\right) + \left(8 + 4x^2\right) \times xv = x^4\)	M1
\(4x^3\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4x^3v = x^4\)	M1
\(4\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4v = x\) *	A1
	(6)

Notes

M1 for attempting to differentiate \(y = xv\) to get \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) – product rule must be used

M1 for differentiating their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain an expression for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) – product rule must be used

A1 for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{\mathrm{d}v}{\mathrm{d}x} + \dfrac{\mathrm{d}v}{\mathrm{d}x} + x\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2}\)

M1 for substituting their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and \(y = xv\) in the original equation to obtain a differential equation in \(v\) and \(x\)

M1 for collecting the terms to have at most a 4 term equation – 4 terms only if a previous error causes \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) to be included, otherwise 3 terms

A1cao and cso for \(4\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4v = x\) *

Alternative for (a)

Scheme	Marks
\(v = \dfrac{y}{x}\)
\(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{1}{x} - y \times \dfrac{1}{x^2}\)	M1
\(\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} = \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \times \dfrac{1}{x} - \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{1}{x^2} - \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{1}{x^2} + 2y \times \dfrac{1}{x^3}\)	M1A1
\(x^3\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} = x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\)	M1
\(4x^3\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4x^3v = 4x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 8x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 8y + 4x^2y = x^4\)	M1
\(4\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4v = x\) *	A1

M1 for writing \(v = \dfrac{y}{x}\) and attempting to differentiate by quotient or product rule to get \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\)

M1 for differentiating their \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) to obtain an expression for \(\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2}\) – product or quotient rule must be used

A1 for \(\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} = \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \times \dfrac{1}{x} - \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{1}{x^2} - \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{1}{x^2} + 2y \times \dfrac{1}{x^3}\)

M1 for multiplying their \(\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2}\) by \(x^3\)

M1 for multiplying by 4 and adding \(4x^2y\) to each side and equating to \(x^4\) (as rhs is now identical to the original equation.

A1cao and cso for \(4\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4v = x\) *

Mark scheme (b)

Scheme	Marks
\(4\lambda^2 + 4 = 0\)
\(\lambda^2 = -1\) oe	M1A1
\((v =)\ C\cos x + D\sin x\) \(\left(\text{or } (v =)\ A\mathrm{e}^{\mathrm{i}x} + B\mathrm{e}^{-\mathrm{i}x}\right)\)	A1
P.I: Try \(v = kx\ (+l)\)
\(\dfrac{\mathrm{d}v}{\mathrm{d}x} = k\quad \dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} = 0\)	M1
\(4 \times 0 + 4\left(kx\ (+l)\right) = x\)	M1dep
\(k = \dfrac{1}{4}\quad (l = 0)\)
\(v = C\cos x + D\sin x + \dfrac{1}{4}x\) \(\left(\text{or } v = A\mathrm{e}^{\mathrm{i}x} + B\mathrm{e}^{-\mathrm{i}x} + \dfrac{1}{4}x\right)\)	A1
	(6)

Notes

M1 for forming the auxiliary equation and attempting to solve

A1 for \(\lambda^2 = -1\) oe

A1 for the complementary function in either form. Award for a correct CF even if \(\lambda = \mathrm{i}\) only is shown.

M1 for trying one of \(v = kx,\ k \neq 1\) or \(v = kx + l\) and \(v = mx^2 + kx + l\) as a PI and obtaining \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2}\)

M1dep for substituting their differentials in the equation \(4\dfrac{\mathrm{d}^2v}{\mathrm{d}x^2} + 4v = x\). Award M0 if the original equation is used. Dep on 2nd M mark of (b)

A1cao for obtaining the correct result (either form)

Mark scheme (c)

Scheme	Marks
\(y = x\left(C\cos x + D\sin x + \dfrac{1}{4}x\right)\) \(\left(\text{or } y = x\left(A\mathrm{e}^{\mathrm{i}x} + B\mathrm{e}^{-\mathrm{i}x} + \dfrac{1}{4}x\right)\right)\)	B1ft
	(1)
	(13 marks)

Notes

B1ft for reversing the substitution to get \(y = x\left(C\cos x + D\sin x + \dfrac{1}{4}x\right)\) \(\left(\text{or } y = x\left(A\mathrm{e}^{\mathrm{i}x} + B\mathrm{e}^{-\mathrm{i}x} + \dfrac{1}{4}x\right)\right)\) follow through their answer to (b)