## Example 2. Solve the equation

$$
\varphi(x)-\lambda \int_{0}^{1} K(x, t) \varphi(t) d t=\cos \pi x
$$

where

$$
K(x, t)= \begin{cases}(x+1) t, & \text { if } \quad 0 \leqslant x \leqslant t \\ (t+1) x, & \text { if } t \leqslant x \leqslant 1\end{cases}
$$
Solution. Characteristic numbers:

$$
\lambda_{0}=1, \quad \lambda_{n}=-n^{2} \pi^{2} \quad(n=1,2, \ldots)
$$

The corresponding eigenfunctions:

$$
\varphi_{0}(x)=e^{x}, \quad \varphi_{n}(x)=\sin n \pi x+n \pi \cos n \pi x \quad(n=1,2, \ldots)
$$

If $\lambda \neq 1$ and $\lambda \neq-n^{2} \pi^{2}$, then the solution of the given equation will have the form

$$
\varphi(x)=\cos \pi x-\lambda\left[\frac{a_{0} e^{x}}{\lambda-1}+\sum_{n=1}^{\infty} \frac{a_{n}}{\lambda+n^{2} \pi^{2}}(\sin n \pi x+n \pi \cos n \pi x)\right]
$$

and since

$$
\begin{aligned}
& a_{0}=\int_{0}^{1} e^{x} \cos \pi x d x=-\frac{1+e}{1+\pi^{2}} \\
& a_{n}=\int_{0}^{1} \cos \pi x(\sin n \pi x+n \pi \cos n \pi x) d x= \begin{cases}0, & n \neq 1 \\
\frac{\pi}{2}, & n=1\end{cases}
\end{aligned}
$$

then

$$
\varphi(x)=\cos \pi x+\lambda\left[\frac{1+e}{1+\pi^{2}} \frac{e^{x}}{\lambda-1}-\frac{\pi}{2\left(\lambda+\pi^{2}\right)}(\sin \pi x+\pi \cos \pi x)\right]
$$

When $\lambda=1$ and $\lambda=-\pi^{2}(n=1)$, the equation has no solutions, because its right-hand side, i.e., the function $\cos \pi x$, is not orthogonal to the corresponding eigenfunctions

$$
\varphi_{0}(x)=e^{x}, \quad \varphi_{1}(x)=\sin \pi x+\pi \cos \pi x
$$

If, however, $\lambda=-n^{2} \pi^{2}$, where $n=2,3, \ldots$, then the given equation has an infinite set of solutions, which are given by the formula (6):

$$
\begin{aligned}
& \varphi(x)=\cos \pi x+\lambda\left[\frac{1+e}{1+\pi^{2}} \frac{e^{x}}{\lambda-1}-\frac{\pi}{2\left(\lambda+\pi^{2}\right)}(\sin \pi x+\pi \cos \pi x)\right]+ \\
&+C(\sin n \pi x+n \pi \cos n \pi x)
\end{aligned}
$$

where $C$ is an arbitrary constant.

In some cases, a non-homogeneous symmetric integral equation can be reduced to a non-homogeneous boundary value problem. This can be done when the kernel $K(x, t)$ of the integral equation is the Green's function of some linear differential operator. Let's show on an example how this is done.