FP2 June 2006 Q1
1. Given that \(3x\sin 2x\) is a particular integral of the differential equation \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4y = k\cos 2x, \qquad \text{where } k \text{ is a constant,}\]
(a) calculate the value of \(k\), (4)
(b) find the particular solution of the differential equation for which at \(x = 0\), \(y = 2\), and for which at \(x = \dfrac{\pi}{4}\), \(y = \dfrac{\pi}{2}\). (4)
| Scheme | Marks |
|---|---|
| \(y' = 3\sin 2x + 6x\cos 2x\) | M1 |
| \(y'' = 12\cos 2x - 12x\sin 2x\) | A1 |
| Substituting \(12\cos 2x - 12x\sin 2x + 12x\sin 2x = k\cos 2x\) | M1 |
| \(k = 12\) | A1 |
| (4) |
Notes
(corrected from the printed mark scheme: the substitution line is printed with \(12x\sin 2\) for the last \(12x\sin 2x\))
| Scheme | Marks |
|---|---|
| General solution is \(y = A\cos 2x + B\sin 2x + 3x\sin 2x\) | B1 |
| \((0, 2) \Rightarrow A = 2\) | B1 |
| \(\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right) \Rightarrow \dfrac{\pi}{2} = B + \dfrac{3\pi}{4} \Rightarrow B = -\dfrac{\pi}{4}\) | M1 |
| \(y = 2\cos 2x - \dfrac{\pi}{4}\sin 2x + 3x\sin 2x\) Needs \(y = \ldots\) | A1 |
| (4) | |
| (8 marks) |