 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/15/b/iii/solution

A = 1

. The original first equation gives

u^{'} (0) = - 3 u (0) + v (0) = - 2

, whereas the general solution gives

u^{'} (0) = B - A

. Thus

B = - 1

and

u (t) = e^{- t} (cos t - sin t) .

The corresponding

v = e^{- t} (cos t - 3 sin t)

has

v (0) = 1

, checking the other initial condition.

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