Discussion
The resolvent kernel \(R(x, t; \lambda)\) for this integral equation is
Answer: A
The resolvent kernel is \(\frac{xe^{t}}{1 - \lambda}\).
The kernel \(K(x, t) = xe^{t}\) is separable. Its iterated kernels are all equal to it, because
\[ \begin{aligned} K_{2}(x, t) &= \int_{0}^{1} xe^{s}\, se^{t}\, ds \\ &= xe^{t}\int_{0}^{1} se^{s}\, ds = xe^{t}\cdot 1. \end{aligned} \]
The resolvent is \(R = \sum_{n \ge 1}\lambda^{n-1}K_{n}\), which here is \(xe^{t}(1 + \lambda + \lambda^{2} + \cdots)\), so
\[ R(x, t; \lambda) = \frac{xe^{t}}{1 - \lambda} \]
for \(|\lambda| \lt 1\).
The key fact is \(\int_{0}^{1}se^{s}\,ds = [se^{s} - e^{s}]_{0}^{1} = 1\). Because every iterated kernel equals \(K\), the series is geometric with ratio \(\lambda\), not \(-\lambda\) or \(\pm\lambda^{2}\), so the denominators \(1 + \lambda\), \(1 + \lambda^{2}\) and \(1 - \lambda^{2}\) cannot arise.
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