 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/a/i/solution

m^{2} - 4 m + 13 = 0, m = 2 \pm 3 i .

y = e^{2 x} (A cos 3 x + B sin 3 x)

. The first boundary value gives

A = π

. At

x = - π /2

cos (- 3 π /2) = 0

and

sin (- 3 π /2) = 1

, so

1 = e^{- π} B

. Hence

y (x) = e^{2 x} (π cos 3 x + e^{π} sin 3 x) .

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