Discussion
Iterated kernel \(K_{m}(t, s)\) of the integral equation
\[ u(t) = 1 + \lambda\int_{0}^{t} e^{t - s}u(s)\,ds \]
is :
Answer: C
The iterated kernel is \(K_{m}(t, s) = \dfrac{(t - s)^{m - 1}}{(m - 1)!}e^{t - s}\).
Here \(K_{1}(t, s) = e^{t - s}\), and \(K_{m}(t, s) = \int_{s}^{t} K(t, r)K_{m-1}(r, s)\,dr\). The exponentials combine to \(e^{t - s}\) inside every integral:
\[ \begin{aligned} K_{2} &= \int_{s}^{t} e^{t - r}e^{r - s}dr = (t - s)e^{t - s} \\ K_{3} &= e^{t - s}\int_{s}^{t}(r - s)dr = \tfrac{(t - s)^{2}}{2}e^{t - s} \end{aligned} \]
By induction \(K_{m} = \tfrac{(t - s)^{m-1}}{(m - 1)!}e^{t - s}\). Option A drops the exponential, option B drops the power, and option D is only \(K_{2}\).
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