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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/15/a/i/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 15 / a / i / Solution by

Codex  0 2026-10-18

The characteristic equation has roots

- 2, - 6

, and a trial multiple of

e^{- 4 t}

gives the particular solution

- 3 e^{- 4 t}

. Thus

f (t) = A e^{- 2 t} + B e^{- 6 t} - 3 e^{- 4 t} .

The first two conditions give

A + B = 3

and

- 2 A - 6 B + 12 = 0

, hence

A = B = 3/2

. At

t_{*} = lo g 2

, the exponentials

e^{- 2 t_{*}}, e^{- 4 t_{*}}, e^{- 6 t_{*}}

equal

1/2, 1/4, 1/8

. Therefore

f (t_{*}) = \frac{3}{4} + \frac{3}{16} - \frac{3}{4} = \frac{3}{16} \neq = 0.

No solution satisfies all three conditions. The first two initial data already determine a unique solution of the second-order linear differential equation; the extra condition is inconsistent with it.

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