Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 2 / 7A / b

Solution

For

t < 0

both inputs vanish. Eliminating

y

gives

x^{''} + 4 x = 0

, and the data at

t = - π

give

x (t) = cos 2 t .

For all

t

, differentiating the first equation and using the second gives

x^{''} + 4 x = δ (t) - 2 H (t - π) .

Thus

x

is continuous at zero while

x^{'}

jumps by one. For

t > 0

the resulting solution is

x (t) = cos 2 t + \frac{1}{2} sin 2 t + \frac{1}{2} H (t - π) (cos 2 t - 1) .

Since

cos 2 t = 1 - 2 sin^{2} t

, this is

x (t) = \frac{1}{2} sin 2 t + 1 + q (t) sin^{2} t,

where

q (t) = {- 2, - 3, 0 < t < π, t \geq π .

Thus $a = 1/2$ , $b = 1$ , and the sketch of $q$ is a downward unit step at $t = π$ .

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