Example 1. Solve the equation

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

where

$$
K(x, t)= \begin{cases}x(t-1), & \text { if } 0 \leqslant x \leqslant t \\ t(x-1), & \text { if } t \leqslant x \leqslant 1\end{cases}
$$
Solution. The eigenvalues and the corresponding eigenfunctions are given by

$$
\lambda_{n}=-\pi^{2} n^{2} ; \quad \varphi_{n}(x)=\sin \pi n x, \quad n=1,2, \ldots .
$$

If $\lambda \neq \lambda_{n}$, then the solution to equation (7) is

$$
\varphi(x)=x-\lambda \sum_{n=1}^{\infty} \frac{a_{n}}{\lambda+n^{2} \pi^{2}} \sin n \pi x
$$

We find the Fourier coefficients $a_{n}$ of the right-hand side of the equation:

$$
a_{n}=\int_{0}^{1} x \sin n \pi x d x=\int_{0}^{1} x d\left(-\frac{\cos n \pi x}{n \pi}\right)=\frac{(-1)^{n+1}}{n \pi}
$$

Substituting into (8), we get

$$
\varphi(x)=x-\frac{\lambda}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n\left(\lambda+n^{2} \pi^{2}\right)} \sin n \pi x
$$

When $\lambda=-n^{2} \pi^{2}$, equation (7) has no solutions, since

$$
a_{n}=\frac{(-1)^{n+1}}{n \pi} \neq 0
$$