Example 1. Solve the system

$$
\left\{\begin{array}{l}
y_{1}^{\prime}=y_{2}+y_{3} \\
y_{2}^{\prime}=y_{1}+y_{3} \\
y_{3}^{\prime}=y_{1}+y_{2}
\end{array}\right.
$$

Find the particular solution if $x=0, y_{1}=1, y_{2}=2, y_{3}=0$.
Solution. We are dealing with a normal and linear system. Let's use the scheme of point $2^{\circ}$.
l) $y_{1}^{\prime \prime}=y_{2}^{\prime}+y_{3}^{\prime}=2 y_{1}+y_{2}+y_{3}=2 y_{1}+y_{1}^{\prime}$.

We differentiated the first equation (with respect to the variable $x$, considered the argument of all three unknown functions of the system), replaced the right side with the sum of the right sides of the second and third equations, and then replaced the sum $y_{2}+y_{3}$ with the left side of the first equation. We obtained a second-order equation with respect to the unknown function $y_{1}$ (obviously, not all operations provided in point $\left.2^{\circ}\right)$ were used):

$$
y_{1}^{\prime \prime}-y_{1}^{\prime}-2 y_{1}=0
$$

Its general solution: $\left(y_{1}\right)_{o 0}=C_{1} e^{-x}+C_{2} e^{2 x}$, or simply (omitting the index oo) $y_{1}=C_{1} e^{-x}+C_{2} e^{2 x}$.

2) To obtain the other functions, we will try to express them through the already found function and its derivatives. Subtract the first equation from the second equation of the system. We get

$$
y_{2}^{\prime}-y_{1}^{\prime}=y_{1}-y_{2}, \quad \text { or } \quad y_{2}^{\prime}+y_{2}=y_{1}^{\prime}+y_{1}
$$

After substituting $y_{1}=C_{1} e^{-x}+C_{2} e^{2 x}, y_{1}^{\prime}=-C_{1} e^{-x}+2 C_{2} e^{2 x}$, we get a new equation $y_{2}^{\prime}+y_{2}=3 C_{2} e^{2 x}$.

The general solution of this non-homogeneous equation, or first-order linear equation ($y_{2}=u v, u^{\prime} v+v^{\prime} u+u v=3 C_{2} e^{2 x}, u=e^{-x}$, $v=C_{2} e^{3 x}+C_{3}$) is the function ($\left.y_{2}\right)_{00}=C_{3} e^{-x}+C_{2} e^{3 x}$.

3) From the first equation $\left(y_{3}\right)_{00}=\left(y_{1}^{\prime}\right)_{00}-\left(y_{2}\right)_{00}=-\left(C_{1}+C_{3}\right) e^{-x}+$ $+C_{2} e^{2 x}$.

The general solution of the system is found, it contains three arbitrary constants: $y_{1}=C_{1} e^{-x}+C_{2} e^{2 x}, y_{2}=C_{3} e^{-x}+C_{2} e^{2 x}, y_{3}=-\left(C_{1}+\right.$ $\left.+C_{3}\right) e^{-x}+C_{2} e^{2 x}$.

4) The initial conditions $x=0, y_{1}=1, y_{2}=2, y_{3}=0$ lead to the system: $\left\{\begin{array}{l}1=C_{1}+C_{2}, \\ 2=C_{2}+C_{3}, \\ 0=-\left(C_{1}+C_{3}\right)+C_{2} .\end{array} \quad\right.$ From this, $C_{1}=0, \quad C_{2}=1, C_{3}=1$.

Answer. $y_{1}=C_{1} e^{-x}+C_{2} e^{2 x}, y_{2}=C_{3} e^{-x}+C_{2} e^{2 x}, y_{3}=-\left(C_{1}+\right.$ $\left.+C_{3}\right) e^{-x}+C_{2} e^{2 x} ;\left(y_{1}\right)_{4}=e^{2 x},\left(y_{2}\right)_{4}=e^{-x}+e^{2 x},\left(y_{3}\right)_{4}=-e^{-x}+e^{2 x}$.

One of the specific methods for integrating systems of differential equations is the method of finding integrable expressions (combinations).