FP2 June 2008 Q5
5.
(a) Find, in terms of \(k\), the general solution of the differential equation \[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 3x = kt + 5,\] where \(k\) is a constant and \(t > 0\). (7)
For large values of \(t\), this general solution may be approximated by a linear function.
(b) Given that \(k = 6\), find the equation of this linear function. (2)
| Scheme | Marks |
|---|---|
| \(m^2 + 4m + 3 = 0\) \(m = -1,\ m = -3\) | M1A1 |
| C.F. \((x =) A\mathrm{e}^{-t} + B\mathrm{e}^{-3t}\) must be function of \(t\), not \(x\) | A1 |
| P.I. \(x = pt + q\) (or \(x = at^2 + bt + c\)) | B1 |
| \(4p + 3(pt + q) = kt + 5\) \(3p = k\) (Form at least one eqn. in \(p\) and/or \(q\)) | M1 |
| \(4p + 3q = 5\) \(p = \dfrac{k}{3},\ q = \dfrac{5}{3} - \dfrac{4k}{9}\ \left(= \dfrac{15 - 4k}{9}\right)\) | A1 |
| General solution: \(x = A\mathrm{e}^{-t} + B\mathrm{e}^{-3t} + \dfrac{kt}{3} + \dfrac{15 - 4k}{9}\) must include \(x =\) and be function of \(t\) | A1ft |
| (7) |
Notes
M1 for auxiliary equation substantially correct
B1 not awarded for \(x = kt +\) constant
| Scheme | Marks |
|---|---|
| When \(k = 6\), \(x = 2t - 1\) | M1 A1cao |
| (2) | |
| (9 marks) |
Notes
M mark for using \(k = 6\) to derive a linear expression in \(t\).
(cf must have involved negative exponentials only)
so e.g. \(y = 2t - 1\) is M1 A0