Example 1. Find the characteristic numbers and eigenfunctions of the integral equation

$$
\varphi(x)-\lambda \int_{0}^{\pi}\left(\cos ^{2} x \cos 2 t+\cos 3 x \cos ^{3} t\right) \varphi(t) d t=0
$$
Solution. Imesm

$$
\varphi(x)=\lambda \cos ^{2} x \int_{0}^{\pi} \varphi(t) \cos 2 t d t+\lambda \cos 3 x \int_{0}^{\pi} \varphi(t) \cos ^{3} t d t
$$

Introducing the notations

$$
C_{1}=\int_{0}^{\pi} \varphi(t) \cos 2 t d t, \quad C_{2}=\int_{0}^{\pi} \varphi(t) \cos ^{3} t d t
$$

we will have

$$
\varphi(x)=C_{1} \lambda \cos ^{2} x+C_{2} \lambda \cos 3 x
$$

Substituting (6) into (5), we obtain a system of linear homogeneous equations

$$
\left\{\begin{aligned}
C_{1}\left(1-\lambda \int_{0}^{\pi} \cos ^{2} t \cos 2 t d t\right)-C_{2} \lambda \int_{0}^{\pi} \cos 3 t \cos 2 t d t & =0 \\
-C_{1} \lambda \int_{0}^{\pi} \cos ^{5} t d t+C_{2}\left(1-\lambda \int_{0}^{\pi} \cos ^{3} t \cos 3 t d t\right) & =0
\end{aligned}\right.
$$

Since

$$
\begin{aligned}
& \int_{0}^{\pi} \cos ^{2} t \cos 2 t d t=\frac{\pi}{4}, \quad \int_{0}^{\pi} \cos 3 t \cos 2 t d t=0 \\
& \int_{0}^{\pi} \cos ^{5} t d t=0, \quad \int_{0}^{\pi} \cos ^{3} t \cos 3 t d t=\frac{\pi}{8}
\end{aligned}
$$

the system (7) will take the form

$$
\left\{\begin{array}{l}
\left(1-\frac{\lambda \pi}{4}\right) C_{1}=0 \\
\left(1-\frac{\lambda \pi}{8}\right) C_{2}=0
\end{array}\right.
$$

The equation for finding the characteristic numbers:

$$
\left|\begin{array}{cc}
1-\frac{\lambda \pi}{4} & 0 \\
0 & 1-\frac{\lambda \pi}{8}
\end{array}\right|=0
$$

Characteristic numbers: $\lambda_{1}=\frac{4}{\pi}, \lambda_{2}=\frac{8}{\pi}$.

For $\lambda=\frac{4}{\pi}$ the system (8) takes the form

$$
\left\{\begin{array}{l}
0 \cdot C_{1}=0 \\
\frac{1}{2} \cdot C_{2}=0
\end{array}\right.
$$

from which $C_{2}=0, C_{1}$ is arbitrary. The eigenfunction will be $\varphi_{1}(x)=C_{1} \lambda \cos ^{2} x$, and, setting $C_{1} \lambda=1$, we get $\varphi_{1}(x)=\cos ^{2} x$.

For $\lambda=\frac{8}{\pi}$ the system (8) will take the form

$$
\left\{\begin{array}{r}
(-1) \cdot C_{1}=0 \\
0 \cdot C_{2}=0
\end{array}\right.
$$

from which $C_{1}=0, C_{2}$ is arbitrary, and, therefore, the eigenfunction will be $\varphi_{2}(x)=$ $C_{2} \lambda \cos 3 x$, or, setting $C_{2} \lambda=1$, we get $\varphi_{2}(x)=\cos 3 x$.

Thus, the characteristic numbers are:

$$
\lambda_{1}=\frac{4}{\pi}, \quad \lambda_{2}=\frac{8}{\pi}
$$

the corresponding eigenfunctions are:

$$
\varphi_{1}(x)=\cos ^{2} x, \quad \varphi_{2}(x)=\cos 3 x
$$

A homogeneous Fredholm integral equation may generally have no characteristic numbers and eigenfunctions, or it may have no real characteristic numbers and eigenfunctions.