FP2 June 2009 Q3
3. Find the general solution of the differential equation \[\sin x\frac{\mathrm{d}y}{\mathrm{d}x} - y\cos x = \sin 2x\sin x,\] giving your answer in the form \(y = \mathrm{f}(x)\). (8)
| Scheme | Marks |
|---|---|
| \(\sin x\dfrac{\mathrm{d}y}{\mathrm{d}x} - y\cos x = \sin 2x\sin x\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{y\cos x}{\sin x} = \dfrac{\sin 2x\sin x}{\sin x}\) An attempt to divide every term in the differential equation by \(\sin x\). Can be implied. | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{y\cos x}{\sin x} = \sin 2x\) | |
| Integrating factor \(= \mathrm{e}^{\int -\frac{\cos x}{\sin x}\,\mathrm{d}x} = \mathrm{e}^{-\ln\sin x}\) \(\mathrm{e}^{\int \pm\frac{\cos x}{\sin x}(\mathrm{d}x)}\) or \(\mathrm{e}^{\int \pm\text{their }\mathrm{P}(x)(\mathrm{d}x)}\) \(\mathrm{e}^{-\ln\sin x}\) or \(\mathrm{e}^{\ln\operatorname{cosec}x}\) | dM1 A1 aef |
| \(= \dfrac{1}{\sin x}\) \(\frac{1}{\sin x}\) or \((\sin x)^{-1}\) or \(\operatorname{cosec}x\) | A1 aef |
| \(\left(\dfrac{1}{\sin x}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{y\cos x}{\sin^2 x} = \dfrac{\sin 2x}{\sin x}\) | |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{y}{\sin x}\right) = \sin 2x \times \dfrac{1}{\sin x}\) \(\frac{\mathrm{d}}{\mathrm{d}x}(y \times \text{their I.F.}) = \sin 2x \times \text{their I.F}\) | M1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{y}{\sin x}\right) = 2\cos x\) \(\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y}{\sin x}\right) = 2\cos x\) or \(\frac{y}{\sin x} = \int 2\cos x\,(\mathrm{d}x)\) | A1 |
| \(\dfrac{y}{\sin x} = \displaystyle\int 2\cos x\,\mathrm{d}x\) | |
| \(\dfrac{y}{\sin x} = 2\sin x + K\) A credible attempt to integrate the RHS with/without \(+K\) | dddM1 |
| \(y = 2\sin^2 x + K\sin x\) \(y = 2\sin^2 x + K\sin x\) | A1 cao |
| (8 marks) |