Discussion
Eigen values of the homogeneous Fredholm integral equation
\[ x(t) = \lambda \int_{0}^{2\pi} \sin(t + s)\, x(s)\, ds \]
are :
Answer: D
The eigenvalues are 1/π and −1/π.
Expand \(\sin(t + s) = \sin t\cos s + \cos t\sin s\), so \(x(t) = \lambda(c_{1}\sin t + c_{2}\cos t)\) with \(c_{1} = \int_{0}^{2\pi}\cos s\, x(s)\, ds\) and \(c_{2} = \int_{0}^{2\pi}\sin s\, x(s)\, ds\). Using \(\int_{0}^{2\pi}\sin^{2} = \int_{0}^{2\pi}\cos^{2} = \pi\) and \(\int_{0}^{2\pi}\sin\cos = 0\):
\[ \begin{aligned} c_{1} &= \lambda\pi c_{2} \\ c_{2} &= \lambda\pi c_{1} \\ c_{1} &= \lambda^{2}\pi^{2}c_{1} \end{aligned} \]
A non-zero solution needs \(\lambda^{2}\pi^{2} = 1\), so \(\lambda = \pm 1/\pi\). The values \(\pm\pi\) are the reciprocals (the eigenvalues of the integral operator itself), and 1 or \(\pi\) do not satisfy the condition.
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