Example 1. Solve the integral equation

$$
\varphi(x)-\lambda \int_{-\pi}^{\pi}\left(x \cos t+t^{2} \sin x+\cos x \sin t\right) \varphi(t) d t=x
$$
Solution. Let's write the equation in the following form:

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

Introduce the notations:

$$
C_{1}=\int_{-\pi}^{\pi} \varphi(t) \cos t d t ; \quad C_{2}=\int_{-\pi}^{\pi} t^{2} \varphi(t) d t ; \quad C_{3}=\int_{-\pi}^{\pi} \varphi(t) \sin t d t
$$

where $C_{1}, C_{2}, C_{3}$ are unknown constants. Then (9) will take the form

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

Substituting expression (11) into equations (10), we get

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

or

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

Calculating the integrals in these equations, we obtain a system of algebraic equations for finding the unknowns $C_{1}, C_{2}, C_{3}$:

$$
\left\{\begin{array}{l}
C_{1}-\lambda \pi C_{3}=0 \\
C_{2}+4 \lambda \pi C_{3}=0 \\
-2 \lambda \pi C_{1}-\lambda \pi C_{2}+C_{3}=2 \pi
\end{array}\right.
$$

The determinant of this system is

$$
\Delta(\lambda)=\left|\begin{array}{ccc}
1 & 0 & -\pi \lambda \\
0 & 1 & 4 \pi \lambda \\
-2 \lambda \pi & -\lambda \pi & 1
\end{array}\right|=1+2 \lambda^{2} \pi^{2} \neq 0
$$

The system (12) has a unique solution

$$
C_{1}=\frac{2 \lambda \pi^{2}}{1+2 \lambda^{2} \pi^{2}} ; \quad C_{2}=-\frac{8 \lambda \pi^{2}}{1+2 \lambda^{2} \pi^{2}} ; \quad C_{3}=\frac{2 \pi}{1+2 \lambda^{2} \pi^{2}}
$$

Substituting the found values of $C_{1}, C_{2}, C_{3}$ into (11), we get the solution of the given integral equation:

$$
\varphi(x)=\frac{2 \lambda \pi}{1+2 \lambda^{2} \pi^{2}}(\lambda \pi x-4 \lambda \pi \sin x+\cos x)+x
$$

## Problems for Independent Solution

Solve the following integral equations with degenerate kernels:

111. $\varphi(x)-4 \int_{0}^{\pi / 2} \sin ^{2} x \varphi(t) d t=2 x-\pi . \quad$ 112. $\varphi(x)-\int_{-1}^{1} e^{\arcsin x} \varphi(t) d t=\operatorname{tg} x$.
112. $\varphi(x)-\lambda \int_{-\pi / 4}^{\pi / 4} \operatorname{tg} t \varphi(t) d t=\operatorname{ctg} x$.
113. $\varphi(x)-\lambda \int_{0}^{1} \cos (q \ln t) \varphi(t) d t=1$.
114. $\varphi(x)-\lambda \int_{0}^{1} \arccos t \varphi(t) d t=\frac{1}{\sqrt{1-x^{2}}}$.
115. $\varphi(x)-\lambda \int_{0}^{1}\left(\ln \frac{1}{t}\right)^{p} \varphi(t) d t=1 \quad(p>-1)$
116. $\varphi(x)-\lambda \int_{0}^{1}(x \ln t-t \ln x) \varphi(t) d t=\frac{6}{5}(1-4 x)$.
117. $\varphi(x)-\lambda \int_{0}^{\pi / 2} \sin x \cos t \varphi(t\rangle d t=\sin x$. 119. $\varphi(x)-\lambda \int_{0}^{2 \pi}|\pi-t| \sin x \varphi(t) d t=x$.
118. $\varphi(x)-\lambda \int_{0}^{\pi} \sin (x-t) \varphi(t) d t=\cos x$.
119. $\varphi(x)-\lambda \int_{0}^{2 \pi}(\sin x \cos t-\sin 2 x \cos 2 t+\sin 3 x \cos 3 t) \varphi(t) d t=\cos x$.
120. $\varphi(x)-\frac{1}{2} \int_{-1}^{1}\left[x-\frac{1}{2}\left(3 t^{2}-1\right)+\frac{1}{2} t\left(3 x^{2}-1\right)\right] \varphi(t) d t=1$

## § 10. Characteristic Numbers and Eigenfunctions

The homogeneous Fredholm integral equation of the second kind

$$
\varphi(x)-\lambda \int_{a}^{b} K(x, t) \varphi(t) d t=0
$$

always has the obvious solution $\varphi(x) \equiv 0$, which is called the zero (trivial) solution.

The values of the parameter $\lambda$ for which this equation has non-zero solutions $\varphi(x) \not=0$ are called characteristic numbers ${ }^{1)}$ of the equation (1) or the kernel $K(x, t)$, and each non-zero solution of this equation is called an eigenfunction corresponding to the characteristic number $\lambda$.

The number $\lambda=0$ is not a characteristic number, since when $\lambda=0$ it follows from (1) that $\varphi(x) \equiv 0$.