Example 1. Solve the boundary value problem

$$
y^{\prime \prime}-y=0, \quad y^{\prime}(0)=0, \quad y(1)=1
$$
Solution. The general solution of the given equation is

$$
y(x)=C_{1} e^{x}+C_{2} e^{-x}
$$

hence

$$
y^{\prime}(x)=C_{1} e^{x}-C_{2} e^{-x}
$$

Setting $x=0$ in (6) and $x=1$ in (5) and considering the boundary conditions, we obtain the following non-homogeneous linear system for finding the values of the constants $C_{1}$ and $C_{2}$

$$
\left\{\begin{array}{r}
C_{1}-C_{2}=0 \\
C_{1} e+C_{2} e^{-1}=1
\end{array}\right.
$$

The determinant of this system is

$$
\Delta=\left|\begin{array}{cc}
1 & -1 \\
e & e^{-1}
\end{array}\right|=e^{-1}+e=2 \operatorname{ch} 1 \neq 0
$$

therefore, it has a unique solution

$$
C_{1}=\frac{1}{2 \mathrm{ch}}, \quad C_{2}=\frac{1}{2 \mathrm{ch} 1}
$$

Substituting the found values of $C_{1}$ and $C_{2}$ into (5), we obtain the solution of the given boundary value problem

$$
y(x)=\frac{e^{x}+e^{-x}}{2 \operatorname{ch} 1}, \quad \text { or } \quad y(x)=\frac{\operatorname{ch} x}{\operatorname{ch} 1}
$$