Example 1. Find the iterated kernels for the kernel $K(x, t)=x-t$, if $a=0, b=1$.
Solution. Using formulas (3), we find sequentially:

$$
\begin{aligned}
& K_{1}(x, t)=x-t, \\
& K_{2}(x, t)=\int_{0}^{1}(x-s)(s-t) d s=\frac{x+t}{2}-x t-\frac{1}{3} \\
& K_{3}(x, t)=\int_{0}^{1}(x-s)\left(\frac{s+t}{2}-s t-\frac{1}{3}\right) d s=-\frac{x-t}{12} \\
& K_{4}(x, t)=-\frac{1}{12} \int_{0}^{1}(x-s)(s-t) d s=-\frac{1}{12} K_{2}(x, t)=-\frac{1}{12}\left(\frac{x+t}{2}-x t-\frac{1}{3}\right),
\end{aligned}
$$

$$
\begin{aligned}
& K_{5}(x, t)=-\frac{1}{12} \int_{0}^{1}(x-s)\left(\frac{s+t}{2}-s t-\frac{1}{3}\right) d s=-\frac{1}{12} K_{3}(x, t)=\frac{x-t}{12^{2}} \\
& K_{6}(x, t)=\frac{1}{12^{2}} \int_{0}^{1}(x-s)(s-t) d s=\frac{K_{2}(x, t)}{12^{2}}=\frac{1}{12^{2}}\left(\frac{x+t}{2}-x t-\frac{1}{3}\right)
\end{aligned}
$$

From this, it follows that the iterated kernels have the form:

1) for $n=2 k-1$

$$
K_{2 k-1}(x, t)=\frac{(-1)^{k-1}}{12^{k-1}}(x-t)
$$

2) for $n=2 k$

$$
K_{2 k}(x, t)=\frac{(-1)^{k-1}}{12^{k-1}}\left(\frac{x+t}{2}-x t-\frac{1}{3}\right),
$$

where $k=1,2,3, \ldots$.