FP2 June 2014 (R) Q8

EdexcelOld spec14 marksSecond Order Differentials

(a) Show that the substitution \(x = \mathrm{e}^z\) transforms the differential equation \[x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2x\frac{\mathrm{d}y}{\mathrm{d}x} - 2y = 3\ln x, \quad x \gt 0 \qquad \text{(I)}\] into the equation \[\frac{\mathrm{d}^2y}{\mathrm{d}z^2} + \frac{\mathrm{d}y}{\mathrm{d}z} - 2y = 3z \qquad \text{(II)}\] (7)

(b) Find the general solution of the differential equation (II). (6)

(c) Hence obtain the general solution of the differential equation (I) giving your answer in the form \(y = \mathrm{f}(x)\). (1)

Mark scheme (a)

Scheme	Marks
\(x = \mathrm{e}^z\)
\(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \mathrm{e}^z\dfrac{\mathrm{d}z}{\mathrm{d}y}\)	M1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-z}\dfrac{\mathrm{d}y}{\mathrm{d}z}\)	A1
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\mathrm{e}^{-z}\dfrac{\mathrm{d}z}{\mathrm{d}x}\times\dfrac{\mathrm{d}y}{\mathrm{d}z} + \mathrm{e}^{-z}\dfrac{\mathrm{d}^2y}{\mathrm{d}z^2}\times\dfrac{\mathrm{d}z}{\mathrm{d}x} = \dfrac{1}{x^2}\left(-\dfrac{\mathrm{d}y}{\mathrm{d}z} + \dfrac{\mathrm{d}^2y}{\mathrm{d}z^2}\right)\)	M1A1A1
\(x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = 3\ln x\)
\(x^2\left(-\dfrac{1}{x^2}\dfrac{\mathrm{d}y}{\mathrm{d}z} + \dfrac{1}{x^2}\dfrac{\mathrm{d}^2y}{\mathrm{d}z^2}\right) + 2x\times\dfrac{1}{x}\dfrac{\mathrm{d}y}{\mathrm{d}z} - 2y = 3z\)	M1
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}z^2} + \dfrac{\mathrm{d}y}{\mathrm{d}z} - 2y = 3z\)	A1
	(7)

Notes

M1 differentiates \(x = \mathrm{e}^z\) wrt \(y\); chain rule must be used
A1 correct differentiation
M1 differentiates again to obtain \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)
A1A1 one mark for each correct term
M1 substitutes in the given equation
A1cso obtains the required equation
ALT:
Works with \(z = \ln x\); marks awarded as above

Alt

Scheme	Marks
\(z = \ln x\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}z}\times\dfrac{\mathrm{d}z}{\mathrm{d}x} = \dfrac{1}{x}\dfrac{\mathrm{d}y}{\mathrm{d}z}\)	M1A1
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{1}{x^2}\dfrac{\mathrm{d}y}{\mathrm{d}z} + \dfrac{1}{x}\dfrac{\mathrm{d}^2y}{\mathrm{d}z^2}\times\dfrac{\mathrm{d}z}{\mathrm{d}x} = -\dfrac{1}{x^2}\dfrac{\mathrm{d}y}{\mathrm{d}z} + \dfrac{1}{x^2}\dfrac{\mathrm{d}^2y}{\mathrm{d}z^2}\)	M1A1A1
\(x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = 3\ln x\)
\(x^2\left(-\dfrac{1}{x^2}\dfrac{\mathrm{d}y}{\mathrm{d}z} + \dfrac{1}{x^2}\dfrac{\mathrm{d}^2y}{\mathrm{d}z^2}\right) + 2x\times\dfrac{1}{x}\dfrac{\mathrm{d}y}{\mathrm{d}z} - 2y = 3z\)	M1
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}z^2} + \dfrac{\mathrm{d}y}{\mathrm{d}z} - 2y = 3z\)	A1
	(7)

Mark scheme (b)

Scheme	Marks
Aux eqn: \(m^2 + m - 2 = 0\)
\((m + 2)(m - 1) = 0\)
\(m = -2,\ 1\)	M1A1
CF: \(y = A\mathrm{e}^{-2z} + B\mathrm{e}^z\)	A1
PI: Try \(y = az + b\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}z} = a \quad \dfrac{\mathrm{d}^2y}{\mathrm{d}z^2} = 0\)
\(a - 2(az + b) = 3z\)
\(a = -\dfrac{3}{2},\quad b = -\dfrac{3}{4}\)	M1
Complete soln: \(y = A\mathrm{e}^{-2z} + B\mathrm{e}^z - \dfrac{3}{2}z - \dfrac{3}{4}\)	A1A1
	(6)

Notes

M1 forms and solves the auxiliary equation
A1 both values for \(m\) correct - may be implied by their CF
A1 correct CF
M1 tries a suitable expression for the PF and obtains values for the constants in the PF
A1A1 shows the complete solution; one mark for each correct term in the PF

Mark scheme (c)

Scheme	Marks
\(y = Ax^{-2} + Bx - \dfrac{3}{2}\ln x - \dfrac{3}{4}\)	B1 ft
	(1)
	(14 marks)

Notes

B1ft reverses the substitution to obtain the solution in the form \(y = \ldots\)
Follow through their complete solution from (b)