ordinary differential equations – NST First Year Maths

2019 Paper 2 Question 15

(b)

Using dot notation for the time derivatives

$\begin{aligned} \dot{u} &= -3u + v \quad(1) \\ \dot{v} &= -5u + v \quad(2) \end{aligned}$

(i)

We need to eliminate $v$ and it’s derivatives. Differentiate (1)

$\ddot{u} = -3 \dot{u} + \dot{v}$

Use (2) to eliminate $\dot{v}$

$\ddot{u} = -3 \dot{u} + -5u + v$

From (1)

$v = \dot{u} + 3u$

and using this to eliminate $v$

$\ddot{u} = -3 \dot{u} + -5u + \dot{u} + 3u$

Hence

$\ddot{u} + 2\dot{u} + 2u = 0$