Example. Solve the equation

$$
\varphi(x)-\lambda \int_{0}^{\pi}\left(\cos ^{2} x \cos 2 t+\cos ^{3} t \cos 3 x\right) \varphi(t) d t=0
$$
Solution. The characteristic numbers of the given equation are $\lambda_{1}=\frac{4}{\pi}$, $\lambda_{2}=\frac{8}{\pi}$, and the corresponding eigenfunctions are

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

The general solution of the equation is

$$
\begin{array}{ll}
\varphi(x)=C \cos ^{2} x, & \text { if } \lambda=\frac{4}{\pi} \\
\varphi(x)=C \cos 3 x, & \text { if } \lambda=\frac{8}{\pi} \\
\varphi(x)=0, & \text { if } \lambda \neq \frac{4}{\pi}, \lambda \neq \frac{8}{\pi},
\end{array}
$$

where $C$ is an arbitrary constant. The last, zero, solution is obtained from the general solutions when $C=0$.

## Problems for Independent Solution

Solve the following homogeneous integral equations:

160. $\varphi(x)-\lambda \int_{0}^{\pi} \cos (x+t) \varphi(t) d t=0$. 161. $\varphi(x)-\lambda \int_{0}^{1} \arccos x \varphi(t) d t=0$.
161. $\varphi(x)-2 \int_{0}^{\pi / 4} \frac{\varphi(t)}{1+\cos 2 t} d t=0$.
162. $\varphi(x)-\frac{1}{4} \int_{-2}^{2}|x| \varphi(t) d t=0$
163. $\varphi(x)+6 \int_{0}^{1}\left(x^{2}-2 x t\right) \varphi(t) d t=0$.

## § 12. Non-homogeneous Symmetric Equations

A non-homogeneous integral equation of Fredholm of the 2nd kind

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

is called symmetric if its kernel $K(x, t)$ is symmetric: $K(x, t) \equiv$ $K(t, x)$.

If $f(x)$ is continuous and the parameter $\lambda$ does not coincide with the characteristic numbers $\lambda_{n}(n=1,2, \ldots)$ of the corresponding homogeneous integral equation

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

then equation (1) has a unique continuous solution, which is given by the formula

$$
\varphi(x)=f(x)-\lambda \sum_{n=1}^{\infty} \frac{a_{n}}{\lambda-\lambda_{n}} \varphi_{n}(x)
$$

where $\varphi_{n}(x)$ are the eigenfunctions of equation (2),

$$
a_{n}=\int_{a}^{b} f(x) \varphi_{n}(x) d x
$$

and the series in the right-hand side of formula (3) converges absolutely and uniformly in the square $a \leqslant x, t \leqslant b$.

If, however, the parameter $\lambda$ coincides with one of the characteristic numbers, say $\lambda=\lambda_{k}$, of rank $q$ (multiplicity of the number $\lambda_{k}$), then equation (1), generally speaking, has no solutions. Solutions exist if and only if the $q$ conditions

$$
\int_{a}^{b} f(x) \varphi_{m}(x) d x=0 \quad(m=1,2, \ldots, q)
$$

are satisfied, i.e., when the function $f(x)$ is orthogonal to all eigenfunctions belonging to the characteristic number $\lambda_{k}$. In this case, equation (1) has an infinite set of solutions, which contain $q$ arbitrary constants and are given by the formula

$$
\begin{aligned}
\varphi(x)=f(x)-\lambda \sum_{n=q+1}^{\infty} & \frac{a_{n}}{\lambda-\lambda_{n}} \varphi_{n}(x)+ \\
+ & C_{1} \varphi_{1}(x)+C_{2} \varphi_{2}(x)+\ldots+C_{q} \varphi_{q}(x)
\end{aligned}
$$

where $C_{1}, C_{2}, \ldots, C_{q}$ are arbitrary constants.

In the case of a degenerate kernel

$$
K(x, t)=\sum_{k=1}^{m} a_{k}(x) b_{k}(t)
$$

formulas (3) and (6) will contain finite sums instead of series in their right-hand sides.

When the right-hand side of equation (1), i.e., the function $f(x)$, is orthogonal to all eigenfunctions $\varphi_{n}(x)$ of equation (2), then the solution of equation (1) is the function itself: $\varphi(x)=f(x)$.