Discussion
Eigenvalues corresponding to the following Sturm-Liouville problem :
\[ \frac{d^{2}X}{dt^{2}} + \lambda X = 0,\ X(0) = 0,\ X'(1) = 0 \]
are \(\lambda_{n}\), \(n \in \mathbf{N}\), where \(\lambda_{n} =\)
Answer: A
The eigenvalues are \(\lambda_{n} = \left[\dfrac{(2n - 1)\pi}{2}\right]^{2}\).
For \(\lambda \le 0\) the boundary conditions force \(X = 0\). For \(\lambda = k^{2} \gt 0\), \(X = A\cos kt + B\sin kt\). \(X(0) = 0\) gives \(A = 0\), and \(X'(1) = Bk\cos k = 0\) gives \(\cos k = 0\), so \(k = (2n - 1)\pi/2\), \(n = 1, 2, \ldots\), and
\[ \lambda_{n} = \left(\frac{(2n - 1)\pi}{2}\right)^{2} \]
\((n\pi/2)^{2}\) would include the even multiples, where \(\cos k = \pm 1\), not 0. \((2n\pi)^{2}\) belongs to periodic conditions, and \((n + 1)\pi\) is not even a square.
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