M5 June 2005 Q2
2. At time \(t\) seconds the position vector of a particle \(P\), relative to a fixed origin \(O\), is \(\mathbf{r}\) metres, where \(\mathbf{r}\) satisfies the differential equation
\[\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} + 2\mathbf{r} = 3\mathrm{e}^{-t}\mathbf{j}.\]Given that \(\mathbf{r} = 2\mathbf{i} - \mathbf{j}\) when \(t = 0\), find \(\mathbf{r}\) in terms of \(t\).
| Scheme | Marks |
|---|---|
| I.F. \(= \mathrm{e}^{\int 2\,\mathrm{d}t} = \mathrm{e}^{2t}\) | B1 |
| \(\Rightarrow\ \mathrm{e}^{2t}\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} + 2\mathrm{e}^{2t}\mathbf{r} = 3\mathrm{e}^{t}\mathbf{j}\) | M1 |
| \(\Rightarrow\ \dfrac{\mathrm{d}}{\mathrm{d}t}(\mathbf{r}\mathrm{e}^{2t}) = 3\mathrm{e}^{t}\mathbf{j}\) | |
| \(\mathrm{e}^{2t}\mathbf{r} = 3\mathrm{e}^{t}\mathbf{j}\ (+\mathbf{c})\) | M1 A1 |
| \(t = 0,\ \mathbf{r} = 2\mathbf{i} - \mathbf{j} \;\Rightarrow\; 2\mathbf{i} - \mathbf{j} = 3\mathbf{j} + \mathbf{c}\) | M1 |
| \(2\mathbf{i} - 4\mathbf{j} = \mathbf{c}\) | A1 |
| \(\Rightarrow\ \mathbf{r} = 3\mathrm{e}^{-t}\mathbf{j} + (2\mathbf{i} - 4\mathbf{j})\mathrm{e}^{-2t}\) | A1 |
| (7 marks) |