 Solution

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/ii/solution

Use the same general solution. The first two conditions now give

A + B = 3

and

- 2 A - 6 B + 12 = - 2

, so

A = 1

B = 2

. Hence

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

t = lo g 2

, its value is

1/2 - 3/4 + 2/8 = 0

, so the third condition is also satisfied. This is the unique consistent solution.

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