4 The motion of a particle is modelled by the differential equation $$v \frac { \mathrm {~d} v } { \mathrm {~d} x } + 4 x = 0$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).

1. Solve the differential equation to show that \(v ^ { 2 } = 20 - 4 x ^ { 2 }\). Now consider motion for which \(x = \cos 2 t + 2 \sin 2 t\), where \(x\) is the displacement from a fixed point at time \(t\).
2. Verify that, when \(t = 0 , x = 1\). Use the fact that \(v = \frac { \mathrm { d } x } { \mathrm {~d} t }\) to verify that when \(t = 0 , v = 4\).
3. Express \(x\) in the form \(R \cos ( 2 t - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v ^ { 2 } = 20 - 4 x ^ { 2 }\).
4. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value.