FP2 June 2011 Q8
8. The differential equation \[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 6\frac{\mathrm{d}x}{\mathrm{d}t} + 9x = \cos 3t, \quad t \geqslant 0\] describes the motion of a particle along the \(x\)-axis.
On the graph of the particular solution defined in part (b), the first turning point for \(t > 30\) is the point \(A\).
| Scheme | Marks |
|---|---|
| \(m^2 + 6m + 9 = 0 \qquad m = -3\) | M1 |
| C.F. \(\quad x = (A + Bt)\mathrm{e}^{-3t}\) | A1 |
| P.I. \(\quad x = P\cos 3t + Q\sin 3t\) | B1 |
| \(\dot{x} = -3P\sin 3t + 3Q\cos 3t\) \(\ddot{x} = -9P\cos 3t - 9Q\sin 3t\) | M1 |
| \((-9P\cos 3t - 9Q\sin 3t) + 6(-3P\sin 3t + 3Q\cos 3t) + 9(P\cos 3t + Q\sin 3t) = \cos 3t\) | M1 |
| \(-9P + 18Q + 9P = 1\) and \(-9Q - 18P + 9Q = 0\) | M1 |
| \(P = 0\) and \(Q = \dfrac{1}{18}\) | A1 |
| \(x = (A + Bt)\mathrm{e}^{-3t} + \dfrac{1}{18}\sin 3t\) | A1ft |
| (8) |
Notes
1st M1 Form auxiliary equation and correct attempt to solve. Can be implied from correct exponential.
2nd M1 for attempt to differentiate PI twice
3rd M1 for substituting their expression into differential equation
4th M1 for substitution of both boundary values
(The printed scheme’s substitution line is cut off at “\(= \cos\)”; it ends \(= \cos 3t\).)
| Scheme | Marks |
|---|---|
| \(t = 0: \quad x = A = \dfrac{1}{2}\) | B1 |
| \(\dot{x} = -3(A + Bt)\mathrm{e}^{-3t} + B\mathrm{e}^{-3t} + \dfrac{3}{18}\cos 3t\) | M1 |
| \(t = 0: \quad \dot{x} = -3A + B + \dfrac{1}{6} = 0 \qquad B = \dfrac{4}{3}\) | M1 A1 |
| \(x = \left(\dfrac{1}{2} + \dfrac{4t}{3}\right)\mathrm{e}^{-3t} + \dfrac{1}{18}\sin 3t\) | A1 |
| (5) |
Notes
1st M1 for correct attempt to differentiate their answer to part (a)
2nd M1 for substituting boundary value
| Scheme | Marks |
|---|---|
| \(t \approx \dfrac{59\pi}{6}\ (\approx 30.9)\) | B1 |
| \(x \approx -\dfrac{1}{18}\) | B1ft |
| (2) | |
| (15 marks) |