 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1/16/c/solution

C cos 2 x + D sin 2 x

(sin x + cos x) /3

, since applying

d^{2} / d x^{2} + 4

to either sine or cosine multiplies it by

3

. Thus

y = C cos 2 x + D sin 2 x + \frac{1}{3} (sin x + cos x) .

At zero,

C + 1/3 = 3

C = 8/3

. At

x = π /4

, the cosine term vanishes and the particular term is

2 /3

, so

D = - 2 /3

. Consequently

y (x) = \frac{8 cos 2 x - 2 sin 2 x + sin x + cos x}{3} .

The two boundary conditions determine the two constants uniquely because their evaluation matrix on

cos 2 x, sin 2 x

is the identity.

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