FP2 June 2005 Q7
7.
(a) Find the general solution of the differential equation \[2\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 5\frac{\mathrm{d}x}{\mathrm{d}t} + 2x = 2t + 9.\] (6)
(b) Find the particular solution of this differential equation for which \(x = 3\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -1\) when \(t = 0\). (4)
The particular solution in part (b) is used to model the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds \((t \geqslant 0)\), \(P\) is \(x\) metres from the origin \(O\).
(c) Show that the minimum distance between \(O\) and \(P\) is \(\tfrac{1}{2}(5 + \ln 2)\) m and justify that the distance is a minimum. (4)
| Scheme | Marks |
|---|---|
| \(2m^2 + 5m + 2 = 0\) Attempt aux eqn \(\to m =\) | M1 |
| \(\Rightarrow m = -\dfrac{1}{2}, -2\) | |
| \(\therefore x_{\text{CF}} = A\mathrm{e}^{-2t} + B\mathrm{e}^{-\frac{1}{2}t}\) C.F. | A1 |
| Particular Integral: \(x = pt + q\) P.I. | B1 |
| \(\dot{x} = p,\ \ddot{x} = 0\) and sub. | M1 |
| \(\Rightarrow 5p + 2q + 2pt = 2t + 9 \to \underline{p = 1, q = 2}\) | A1 |
| General solution \(x = \underline{A\mathrm{e}^{-2t} + B\mathrm{e}^{-\frac{1}{2}t}}\ \underline{{} + t + 2}\) | A1ft (ft ms, p.q) |
| (6) |
Notes
(corrected from the printed mark scheme: the right-hand side of the equation for \(p\) and \(q\) is printed as \(2t + q\); it should be \(2t + 9\))
| Scheme | Marks |
|---|---|
| \(x = 3, t = 0 \Rightarrow 3 = A + B + 2\) (or \(A + B = 1\)) | M1 |
| \(\dot{x} = -2A\mathrm{e}^{-2t} - \dfrac{1}{2}B\mathrm{e}^{-\frac{1}{2}t} + 1\) Attempt \(\dot{x}\) | M1 |
| \(\dot{x} = -1, t = 0 \Rightarrow -1 = -2A - \dfrac{1}{2}B + 1\) (or \(4A + B = 4\)) 2 correct eqns | A1 |
| Solving \(\to A = 1, B = 0\) and \(\underline{x = \mathrm{e}^{-2t} + t + 2}\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\dot{x} = -2\mathrm{e}^{-2t} + 1 = 0\) \(\dot{x} = 0\) | M1 |
| \(\Rightarrow t = \dfrac{1}{2}\ln 2\) | A1 |
| \(\ddot{x} = 4\mathrm{e}^{-2t} \gt 0\ (\forall t)\ \therefore \text{min}\) | M1 |
| Min \(x = \mathrm{e}^{-\ln 2} + \dfrac{1}{2}\ln 2 + 2\) \(= \dfrac{1}{2} + \dfrac{1}{2}\ln 2 + 2\) \(= \dfrac{1}{2}\,\underline{(5 + \ln 2)}\ (*)\) | A1 c.s.o. |
| (4) | |
| (14 marks) |