Discussion
Eigen value and eigen function of the homogeneous integral equation
\(u(t) = \lambda\displaystyle\int_{0}^{1} e^{t+s}\, u(s)ds\)
is :
Answer: B
The eigenvalue and eigenfunction are \(\lambda = \dfrac{2}{e^{2} - 1}\), \(u(t) = e^{t}\).
The kernel \(e^{t+s} = e^{t}e^{s}\) is separable, so \(u(t) = \lambda e^{t}C\) with \(C = \int_{0}^{1} e^{s}u(s)\,ds\). Substituting \(u = \lambda Ce^{t}\):
\[ \begin{aligned} C &= \lambda C\int_{0}^{1} e^{2s}\,ds = \lambda C\,\frac{e^{2} - 1}{2} \\ \lambda &= \frac{2}{e^{2} - 1} \end{aligned} \]
for a non-zero C. The eigenfunction is a multiple of \(e^{t}\), not \(e^{t^{2}}\), and the reciprocal value \((e^{2} - 1)/2\) is the integral itself, not the eigenvalue.
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