 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2012/8/b/solution

For this forcing the integral in part (a) simplifies to

y_{p} (x) = e^{- 2 x} \int_{0}^{x} sin (x - t) d t = e^{- 2 x} (1 - cos x), C = - \int_{0}^{π /2} cos t d t = - 1.

Consequently the unique boundary value problem solution is

y (x) = e^{- 2 x} (1 - sin x - cos x) .

It vanishes at zero and at

π /2

. To check the differential equation directly, set

y = e^{- 2 x} w

. Then

L y = e^{- 2 x} (w^{''} + w)

. For

w = 1 - sin x - cos x

, one has

w^{''} + w = 1

, as required.

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