M5 June 2011 Q2
2. A particle \(P\) moves in the \(x\)-\(y\) plane so that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation
\[\frac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2} - 4\mathbf{r} = -3\mathrm{e}^{t}\mathbf{j}\]When \(t = 0\), the particle is at the origin and is moving with velocity \((2\mathbf{i} + \mathbf{j})\) m s\(^{-1}\).
Find \(\mathbf{r}\) in terms of \(t\). (10)
| Scheme | Marks |
|---|---|
| \(m^2 - 4 = 0 \Rightarrow m = 2\) or \(-2\) | M1 |
| CF is \(\mathbf{r} = \mathbf{A}\mathrm{e}^{2t} + \mathbf{B}\mathrm{e}^{-2t}\) | A1 |
| PI try \(\mathbf{r} = \mathbf{C}\mathrm{e}^{t}\) \(\dot{\mathbf{r}} = \mathbf{C}\mathrm{e}^{t}\) \(\ddot{\mathbf{r}} = \mathbf{C}\mathrm{e}^{t}\) | B1 |
| \(\mathbf{C}\mathrm{e}^{t} - 4\mathbf{C}\mathrm{e}^{t} = -3\mathrm{e}^{t}\mathbf{j}\) | M1 |
| \(\mathbf{C} = \mathbf{j}\) | A1 |
| GS is \(\mathbf{r} = \mathbf{A}\mathrm{e}^{2t} + \mathbf{B}\mathrm{e}^{-2t} + \mathbf{j}\mathrm{e}^{t}\) | A1 |
| \(\mathbf{v} = 2\mathbf{A}\mathrm{e}^{2t} - 2\mathbf{B}\mathrm{e}^{-2t} + \mathbf{j}\mathrm{e}^{t}\) | M1 |
| \(t = 0,\ \mathbf{r} = \mathbf{0},\ \mathbf{v} = 2\mathbf{i} + \mathbf{j}\) | M1 |
| \(\mathbf{0} = \mathbf{A} + \mathbf{B} + \mathbf{j}\) \(2\mathbf{i} + \mathbf{j} = 2\mathbf{A} - 2\mathbf{B} + \mathbf{j}\) | A1 |
| \(\mathbf{i} = \mathbf{A} - \mathbf{B}\) \(\mathbf{A} = \dfrac{1}{2}(\mathbf{i} - \mathbf{j});\ \mathbf{B} = -\dfrac{1}{2}(\mathbf{i} + \mathbf{j})\) | |
| \(\mathbf{r} = \dfrac{1}{2}(\mathbf{i} - \mathbf{j})\mathrm{e}^{2t} - \dfrac{1}{2}(\mathbf{i} + \mathbf{j})\mathrm{e}^{-2t} + \mathbf{j}\mathrm{e}^{t}\) | A1 |
| (10 marks) |