FP2 June 2017 Q5

(a) Find the general solution of the differential equation \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 26\sin 3x\] (8)

(b) Find the particular solution of this differential equation for which \(y = 0\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 0\) (5)

Mark scheme (a)

Scheme	Marks
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 26\sin 3x\)
\(m^2 - 2m = 0 \Rightarrow m = 0,\ 2\) Solves AE	M1
(CF or \(y =\)) \(A + B\mathrm{e}^{2x}\) or \(A\mathrm{e}^0 + B\mathrm{e}^{2x}\) oe Correct CF (CF or \(y =\) not needed)	A1
(PI or \(y =\)) \(a\cos 3x + b\sin 3x\) Correct form for PI (PI or \(y =\) not needed)	B1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3a\sin 3x + 3b\cos 3x,\ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -9a\cos 3x - 9b\sin 3x\) M1: Differentiates twice; change of trig functions needed, \(\pm1\) or \(\pm3\) for coeffs for first derivative, \(\pm1,\ \pm3\) or \(\pm9\) for second derivative (1/3 etc indicates integration) A1: Correct derivatives	M1A1
\(-9a\cos 3x - 9b\sin 3x + 6a\sin 3x - 6b\cos 3x = 26\sin 3x\)
\(\therefore -9a - 6b = 0,\ \ -9b + 6a = 26 \Rightarrow a = \ldots, b = \ldots\) Substitutes and forms simultaneous equations (by equating coeffs) and attempts to solve for \(a\) and \(b\) Depends on the second M mark	dM1
\(a = \dfrac{4}{3},\ b = -2\) Correct \(a\) and \(b\)	A1
\(y = A + B\mathrm{e}^{2x} + \dfrac{4}{3}\cos 3x - 2\sin 3x\) Forms the GS (ft their CF and PI) Must start \(y = \ldots\)	A1ft
	(8)

Scheme	Marks
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 26\sin 3x \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = -\dfrac{26}{3}\cos 3x + c\) M1: Integrates both sides wrt \(x\) A1: Correct expression	M1A1
\(I = \mathrm{e}^{\int -2\,\mathrm{d}x} = \mathrm{e}^{-2x}\) Correct integrating factor	B1
\(y\mathrm{e}^{-2x} = \displaystyle\int \mathrm{e}^{-2x}\left(-\frac{26}{3}\cos 3x + c\right)\mathrm{d}x\) M1: Uses \(yI = \displaystyle\int I\left(-\frac{26}{3}\cos 3x + c\right)\mathrm{d}x\) A1: Correct expression	M1A1
\(= \dfrac{4}{3}\mathrm{e}^{-2x}\cos 3x - 2\mathrm{e}^{-2x}\sin 3x - \dfrac{1}{2}c\mathrm{e}^{-2x} + B\) M1: Integration by parts twice A1: Correct expression	M1A1
\(y = -\dfrac{1}{2}c + B\mathrm{e}^{2x} + \dfrac{4}{3}\cos 3x - 2\sin 3x\) Must start \(y = \ldots\)

Mark scheme (b)

Scheme	Marks
\(0 = A + B + \dfrac{4}{3}\) Substitutes \(x = 0\) and \(y = 0\) into their GS	M1
\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = 2B\mathrm{e}^{2x} - 4\sin 3x - 6\cos 3x \Rightarrow 0 = 2B - 6\) Differentiates and substitutes \(x = 0\) and \(y' = 0\) (change of trig functions needed, \(\pm1\) or \(\pm3\) for coeffs )	M1
\(0 = A + B + \dfrac{4}{3},\ 0 = 2B - 6 \Rightarrow A = \ldots, B = \ldots\) Solves simultaneously to obtain values for \(A\) and \(B\) Depends on the second M mark	dM1
\(A = \dfrac{-13}{3},\quad B = 3\) Correct values	A1
\(y = 3\mathrm{e}^{2x} - \dfrac{13}{3} + \dfrac{4}{3}\cos 3x - 2\sin 3x\) Follow through their GS and \(A\) and \(B\) Must start \(y = \ldots\)	A1ft
	(5)
	(13 marks)