Example 2. Solve the integral equation

$$
\varphi(x)=x+\int_{0}^{x} x t \varphi(t) d t
$$
Solution. Rewrite equation (11) in the following form:

$$
\varphi(x)=x\left(1+\int_{0}^{x} t \varphi(t) d t\right)
$$

and set

$$
y(x)=1+\int_{0}^{x} t \varphi(t) d t
$$

Differentiate the last equality:

$$
y^{\prime}(x)=x \varphi(x)
$$

But since according to (12) and (13)

$$
\varphi(x)=x y(x),
$$

we obtain a differential equation for the function $y(x)$:

$$
y^{\prime}(x)=x^{2} y(x)
$$

The general solution of this equation is

$$
y(x)=C \mathrm{e}^{x^{3} / 3}
$$

Note that due to (13) we have $y(0)=1$ and, consequently, $C=1$. Thus, the solution $\varphi(x)=x y(x)$ of equation (11) is

$$
\varphi(x)=x e^{x^{3} / 3}
$$

## Problems for Independent Solution

Solve the following integral equations by the method of differentiation:

19. $\varphi(x)=x-\int_{0}^{x} e^{x-t} \varphi(t) d t . \quad 20 . \int_{0}^{x} e^{x+t} \varphi(t) d t=x, \quad 21 . \quad \int_{0}^{x} e^{x-t} \varphi(t) d t=x$.
20. $\varphi(x)=2 \int_{0}^{x} \frac{2 t+1}{(2 x+1)^{2}} \varphi(t) d t+1 . \quad$ 23. $\varphi(x)=\int_{0}^{x} \varphi(t) d t+e^{x}$.

## § 3. Resolvent of an Integral Equation of Volterra. Solving an Integral Equation Using the Resolvent

Suppose we have an integral equation of Volterra of the second kind

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

where $K(x, t)$ is a continuous function for $0 \leqslant x \leqslant a, 0 \leqslant t \leqslant x$, and $f(x)$ is continuous for $0 \leqslant x \leqslant a$.

We seek the solution of the integral equation (1) in the form of an infinite power series in powers of $\lambda$:

$$
\varphi(x)=\varphi_{0}(x)+\lambda \varphi_{1}(x)+\lambda^{2} \varphi_{2}(x)+\ldots+\lambda^{n} \varphi_{n}(x)+\ldots
$$

Formally substituting this series into (1), we get

$$
\begin{aligned}
& \varphi_{0}(x)+\lambda \varphi_{1}(x)+\ldots+\lambda^{n} \varphi_{n}(x)+\ldots= \\
& =f(x)+\lambda \int_{0}^{x} K(x, t)\left[\varphi_{0}(t)+\lambda \varphi_{1}(t)+\ldots+\lambda^{n} \varphi_{n}(t)+\ldots\right] d t
\end{aligned}
$$

Comparing the coefficients of the same powers of $\lambda$, we find

$$
\begin{aligned}
\varphi_{0}(x)=f(x), \\
\varphi_{1}(x)=\int_{0}^{x} K(x, t) \varphi_{0}(t) d t=\int_{0}^{x} K(x, t) f(t) d t, \\
\varphi_{2}(x)=\int_{0}^{x} K(x, t) \varphi_{1}(t) d t=\int_{0}^{x} K(x, t) \int_{0}^{t} K\left(t, t_{1}\right) f\left(t_{1}\right) d t_{1} d t,
\end{aligned}
$$

The relations (3) provide a method for sequentially defining the functions $\varphi_{n}(x)$. It can be shown that under the assumptions made about $f(x)$ and $K(x, t)$, the series (2) converges uniformly in $x$ and $\lambda$ for any $\lambda$ and $x \in[0, a]$, and its sum is the unique solution of equation (1).

Furthermore, from (3) it follows:

$$
\begin{aligned}
\varphi_{1}(x) & =\int_{0}^{x} K(x, t) f(t) d t, \\
\varphi_{2}(x) & =\int_{0}^{x} K(x, t)\left[\int_{0}^{t} K\left(t, t_{1}\right) f\left(t_{1}\right) d t_{1}\right] d t= \\
& =\int_{0}^{x} f\left(t_{1}\right) d t_{1} \int_{t_{1}}^{x} K(x, t) K\left(t, t_{1}\right) d t=\int_{0}^{x} K_{2}\left(x, t_{1}\right) f\left(t_{1}\right) d t_{1},
\end{aligned}
$$

where

$$
K_{2}\left(x, t_{1}\right)=\int_{t_{1}}^{x} K(x, t) K\left(t, t_{1}\right) d t
$$

Similarly, it can be established that generally

$$
\varphi_{n}(x)=\int_{0}^{x} K_{n}(x, t) f(t) d t \quad(n=1,2, \ldots)
$$

The functions $K_{n}(x, t)$ are called repeated or iterated kernels. They, as can be easily shown, are defined by the recurrence formulas

$$
\begin{gathered}
K_{1}(x, t)=K(x, t) \\
K_{n+1}(x, t)=\int_{t}^{x} K(x, z) K_{n}(z, t) d z \quad(n=1,2, \ldots)
\end{gathered}
$$

Using (4) and (5), equality (2) can be written as:

$$
\varphi(x)=f(x)+\sum_{\nu=1}^{\infty} \lambda^{\nu} \int_{0}^{x} K_{\nu}(x, t) f(t) d t
$$

The function $R(x, t ; \lambda)$, defined by the series

$$
R(x, t ; \lambda)=\sum_{\nu=0}^{\infty} \lambda^{\nu} K_{\nu+1}(x, t)
$$

is called the resolvent (or resolving kernel) of the integral equation (1). The series (6) converges absolutely and uniformly in the case of a continuous kernel $K(x, t)$.

The repeated kernels, as well as the resolvent, do not depend on the lower limit in the integral equation.

The resolvent $R(x, t ; \lambda)$ satisfies the following functional equation:

$$
R(x, t ; \lambda)=K(x, t)+\lambda \int_{t}^{x} K(x, s) R(s, t ; \lambda) d s
$$

Using the resolvent, the solution of the integral equation (1) can be written as

$$
\varphi(x)=f(x)+\lambda \int_{0}^{x} R(x, t ; \lambda) f(t) d t
$$