 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-4/12a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The Bromwich inversion integral closes to the left for

t > 0

. Equivalently, partial fractions at the two double poles give

\frac{1}{s ^{2} ( s + 2 ) ^{2}} = - \frac{1}{4 s} + \frac{1}{4 s ^{2}} + \frac{1}{4 ( s + 2 )} + \frac{1}{4 ( s + 2 ) ^{2}} .

Taking residues therefore yields

f (t) = \frac{1}{4} (t - 1 + (t + 1) e^{- 2 t}) .

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