FP2 June 2005 Q2
2. Find the general solution of the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\cot 2x = \sin x, \qquad 0 \lt x \lt \frac{\pi}{2},\] giving your answer in the form \(y = \mathrm{f}(x)\).
| Scheme | Marks |
|---|---|
| I.F. \(= \mathrm{e}^{\int 2\cot 2x\,\mathrm{d}x}\); \(= \sin 2x\) | M1 A1 |
| Multiplying throughout by IF. | M1(*) |
| \(y \times (\text{IF}) =\) integral of candidate’s RHS | M1 |
| \(= \displaystyle\int 2\sin^2 x\cos x\,\mathrm{d}x\) or \(\displaystyle\int -\left(\frac{\cos 3x - \cos x}{2}\right)\mathrm{d}x\) [This M gained when in position to complete integration, dep on M(*)] | M1 |
| \(= \dfrac{2}{3}\sin^3 x\,(+C)\) or \(-\dfrac{1}{6}\sin 3x + \dfrac{1}{2}\sin x + c\) | A1 |
| \(y = \dfrac{2\sin^3 x}{3\sin 2x} + \dfrac{C}{\sin 2x}\) or \(-\dfrac{\sin 3x}{6\sin 2x} + \dfrac{\sin x}{2\sin 2x} + \dfrac{c}{\sin 2x}\) or equiv. | A1ft |
| (7 marks) |