 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2008/a/2/b/solution

y^{''} + (1 + λ) y = 0

. If

1 + λ \leq 0

, the linear or hyperbolic branches cannot meet both homogeneous endpoint conditions nontrivially. Otherwise

y = C sin (k x)

k = 1 + λ

, and

y^{'} (1) = C k cos k = 0

gives

k_{n} = (n + 1/2) π

. Therefore

λ_{n} = (n + \frac{1}{2})^{2} π^{2} - 1, ϕ_{n} (x) = 2 sin [(n + \frac{1}{2}) π x], n = 0, 1, \dots .

The factor

2

makes the eigenfunctions unit normalized. All the eigenvalues are positive, in particular none is zero.

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