Discussion
The second smallest eigenvalue of the eigenvalue problem
\[ \begin{aligned} &\frac{d^{2}y}{dx^{2}} + (\lambda - 3)y = 0, \\ &y(0) = y(\pi) = 0, \end{aligned} \]
is
Answer: C
The second smallest eigenvalue is 7.
Write \(\mu = \lambda - 3\). The problem \(y'' + \mu y = 0\), \(y(0) = y(\pi) = 0\) has non-zero solutions only when \(\mu = n^{2}\), \(n = 1, 2, 3, \ldots\), with \(y = \sin nx\). (For \(\mu \le 0\) the only solution meeting both conditions is \(y = 0\).)
\[ \lambda = n^{2} + 3 = 4,\ 7,\ 12,\ \ldots \]
The smallest is 4 and the second smallest is 7.
4 is the smallest, not the second. 3 would need \(\mu = 0\), which gives only the zero solution. 9 is \(n^{2}\) for \(n = 3\) without the shift of 3.
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