Example 2. Find the resolvent of the integral equation

$$
\varphi(x)=f(x)+\int_{0}^{x}(x-t) \varphi(t) d t
$$
Solution. In this case, $K(x, t)=x - t, \lambda=1$, hence, according to (8), $a_{1}(x)=1$, and all other $a_{k}(x)=0$.

Equation (9) in this case has the form

$$
\frac{d^{2} g(x, t ; 1)}{d x^{2}}-g(x, t ; 1)=0
$$

from which

$$
g(x, t ; 1)=g(x, t)=C_{1}(t) e^{x}+C_{2}(t) e^{-x}
$$

Conditions (10) give

$$
\left\{\begin{array}{l}
C_{1}(t) e^{t}+C_{2}(t) e^{-t}=0 \\
C_{1}(t) e^{t}-C_{2}(t) e^{-t}=1
\end{array}\right.
$$

Solving system (12), we find

$$
C_{1}(t)=\frac{1}{2} e^{-t}, \quad C_{2}(t)=-\frac{1}{2} e^{t}
$$

and, therefore;

$$
g(x, t)=\frac{1}{2}\left(e^{x-t}-e^{-(x-t)}\right)=\operatorname{sh}(x-t)
$$

According to (11)

$$
R(x, t ; 1)=[\operatorname{sh}(x-t)]_{x}^{\prime \prime}=\operatorname{sh}(x-t)
$$

## Problems for Independent Solution

Find the resolvents of integral equations with the following kernels ( $\lambda=1$ ):

31. $K(x, t)=2-(x-t)$.
32. $K(x, t)=-2+3(x-t)$.
33. $K(x, t)=2 x$.
34. $K(x, t)=-\frac{4 x-2}{2 x+1}+\frac{8(x-t)}{2 x+1}$.
35. Suppose we have a Volterra integral equation, the kernel of which depends only on the difference of its arguments:

$$
\varphi(x)=f(x)+\int_{0}^{x} K(x-t) \varphi(t) d t \quad(\lambda=1)
$$

Show that for equation (13) all repeated kernels and the resolvent also depend only on the difference $x-t$.