 Solution

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

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

The first three initial values vanish and the third derivative at zero equals one. Transforming the differential equation gives

(s^{4} + 4 s^{3} + 4 s^{2}) F (s) - 1 = 0,

F (s) = 1/ [s^{2} (s + 2)^{2}]

. Hence

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

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