 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/1/b/iii/solution

b_{n} = ⟨ y_{n}, 1 ⟩_{w} = 2 \int_{0}^{1} sin (nπ t) d t = \frac{2 [ 1 - ( - 1 ) ^{n} ]}{nπ} .

Every

λ_{n}

is nonzero, so the solution coefficient is

b_{n} / λ_{n}

. Thus

y (x) = n = 1 \sum \infty \frac{2 [ 1 - ( - 1 ) ^{n} ]}{nπ ( n ^{2} π ^{2} - 1 )} sin (nπ lo g x) = n \geq 1 n odd \sum \frac{4 sin ( nπ lo g x )}{nπ ( n ^{2} π ^{2} - 1 )} .

As an independent check, solving

Y^{''} + Y = - 1

directly with the two endpoint conditions gives

y (x) = \frac{c o s ( l o g x - \frac{1}{2} )}{c o s ( \frac{1}{2} )} - 1.

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