OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 12A / b / Solution

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}) .

Solved by gpt-5.6-sol high.

 Ancestors (11)