Example 1. Find the resolvent of the Volterra integral equation with the kernel $K(x, t) \equiv 1$.
Solution. We have $K_{1}(x, t)=K(x, t)=1$. Further, according to formulas (5)

\[
\begin{aligned}
& K_{2}(x, t)=\int_{t}^{x} K(x, z) K_{1}(z, t) d z=\int_{t}^{x} d z=x-t \\
& K_{3}(x, t)=\int_{t}^{x} 1 \cdot(z-t) d z=\frac{(x-t)^{2}}{2} \\
& K_{4}(x, t)=\int_{t}^{x} 1 \cdot \frac{(z-t)^{2}}{2} d z=\frac{(x-t)^{3}}{3!}
\end{aligned}
\]

\[
K_{n}(x, t)=\int_{t}^{x} 1 \cdot K_{n-1}(z, t) d z=\int_{t}^{x} 1 \cdot \frac{(z-t)^{n-2}}{(n-2)!} d z=\frac{(x-t)^{n-1}}{(n-1)!}
\]

Thus, according to the definition

\[
R(x, t ; \lambda)=\sum_{n=0}^{\infty} \lambda^{n} K_{n+1}(x, t)=\sum_{n=0}^{\infty} \frac{\lambda^{n}(x-t)^{n}}{n!}=e^{\lambda(x-t)}
\]

## Problems for Independent Solution

Find the resolvents for the Volterra integral equations with the following kernels:
24. $K(x, t)=x-t$.
25. $K(x, t)=e^{x-t}$.
26. $K(x, t)=e^{x^{2}-t^{2}}$.

27. $K(x, t)=\frac{1+x^{2}}{1+t^{2}}$.
28. $K(x, t)=\frac{2+\cos x}{2+\cos t}$.
29. $K(x, t)=\frac{\operatorname{ch} x}{\operatorname{ch} t}$.
30. $K(x, t)=a^{x-t}(a>0)$.

Suppose the kernel $K(x, t)$ is a polynomial of degree $(n-1)$ in $t$, so it can be represented as

\[
K(x, t)=a_{0}(x)+a_{1}(x)(x-t)+\ldots+\frac{a_{n-1}(x)}{(n-1)!}(x-t)^{n-1}
\]

where the coefficients $a_{k}(x)$ are continuous on $[0, a]$. If we define the function $g(x, t ; \lambda)$ as the solution to the differential equation

\[
\frac{d^{n} g}{d x^{n}}-\lambda\left[a_{0}(x) \frac{d^{n-1} g}{d x^{n-1}}+a_{1}(x) \frac{d^{n-2} g}{d x^{n-2}}+\ldots+a_{n-1}(x) g\right]=0
\]

satisfying the conditions

\[
\left.g\right|_{x=t}=\left.\frac{d g}{d x}\right|_{x=t}=\ldots=\left.\frac{d^{n-2} g}{d x^{n-2}}\right|_{x=t}=0,\left.\quad \frac{d^{n-1} g}{d x^{n-1}}\right|_{x=t}=1
\]

then the resolvent $R(x, t ; \lambda)$ is defined by the equality

\[
R(x, t ; \lambda)=\frac{1}{\lambda} \frac{d^{n} g(x, t ; \lambda)}{d x^{n}}
\]

Similarly, in the case when

\[
K(x, t)=b_{0}(t)+b_{1}(t)(t-x)+\ldots+\frac{b_{n-1}(t)}{(n-1)!}(t-x)^{n-1}
\]

the resolvent

\[
R(x, t ; \lambda)=-\frac{1}{\lambda} \frac{d^{n} g(t, x ; \lambda)}{d t^{n}}
\]

where $g(x, t ; \lambda)$ is the solution to the equation

\[
\frac{d^{n} g}{d t^{n}}+\lambda\left[b_{0}(t) \frac{d^{n-1} g}{d t^{n-1}}+\ldots+b_{n-1}(t) g\right]=0
\]

satisfying the conditions (10).