Discussion
Eigenvalues of the Fredholm integral equation,
\[ x(t) = 1 + \lambda \int_{0}^{\pi} \cos(t + s)\, x(s)\, ds \]
are :
Answer: A
The eigenvalues are \(2/\pi\) and \(-2/\pi\).
Expand the kernel: \(\cos(t + s) = \cos t\cos s - \sin t\sin s\). For the homogeneous equation, \(x(t) = \lambda(A\cos t - B\sin t)\) with \(A = \int_{0}^{\pi} \cos s\, x(s)\,ds\) and \(B = \int_{0}^{\pi} \sin s\, x(s)\,ds\).
On \([0, \pi]\), \(\int \cos^{2} = \int \sin^{2} = \frac{\pi}{2}\) and \(\int \sin s\cos s\,ds = 0\), so
\[ A = \lambda\frac{\pi}{2}A, \qquad B = -\lambda\frac{\pi}{2}B \]
A non-trivial solution needs \(\lambda = \frac{2}{\pi}\) (with \(x = \cos t\)) or \(\lambda = -\frac{2}{\pi}\) (with \(x = \sin t\)).
The pairs \(\pi, -\pi\), \(1, -1\) and \(\sqrt{2}, -\sqrt{2}\) do not satisfy \(\lambda\frac{\pi}{2} = \pm 1\), so they are not eigenvalues.
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