Example 2. Solve the system

$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=\frac{x_{1}-x_{2}}{x_{3}-t} \\
\frac{d x_{2}}{d t}=\frac{x_{1}-x_{2}}{x_{3}-t} \\
\frac{d x_{3}}{d t}=x_{1}-x_{2}+1
\end{array}\right.
$$
Solution. Subtracting the second equation from the first term by term, we get $\frac{d\left(x_{1}-x_{2}\right)}{d t}=0$, from which the first integral of the system (4)

$$
x_{1}-x_{2}=C_{1}
$$

Substituting (5) into the second and third equations of the system (4), we obtain a system with two unknown functions

$$
\left\{\begin{array}{l}
\frac{d x_{2}}{d t}=\frac{C_{1}}{x_{3}-t} \\
\frac{d x_{3}}{d t}=C_{1}+1
\end{array}\right.
$$

From the second equation of the system (6) we find

$$
x_{3}=\left(C_{1}+1\right) t+C_{2}
$$

Substituting (7) into the first equation of the system (6), we will have

$$
\frac{d x_{2}}{d t}=\frac{C_{1}}{C_{1} t+C_{2}}, \quad x_{2}=\ln \left|C_{1} t+C_{2}\right|+C_{3}
$$

Thus,

$$
x_{1}-x_{2}=C_{1}, \quad x_{2}=\ln \left|C_{1} t+C_{2}\right|+C_{3}, \quad x_{3}=\left(C_{1}+1\right) t+C_{2}
$$

From this we find the general solution of the system (4):

$$
x_{1}=\ln \left|C_{1} t+C_{2}\right|+C_{1}+C_{3}, \quad x_{2}=\ln \left|C_{1} t+C_{2}\right|+C_{3}, \quad x_{3}=\left(C_{1}+1\right) t+C_{2} . \Delta
$$