Example 5. Solve the system of differential equations

$$
\left\{\begin{array}{l}
x^{\prime}=2 x-4 y \\
y^{\prime}=x-3 y
\end{array}\right.
$$
Solution. Let $A=\left(\begin{array}{cc}2 & -4 \\ 1 & -3\end{array}\right)$ be the matrix composed of the coefficients from the right-hand side of the system. We write down the characteristic equation of matrix $A$:

$$
\left|\begin{array}{cc}
2-\lambda & -4 \\
1 & -3-\lambda
\end{array}\right|=-6+3 \lambda-2 \lambda+\lambda^{2}+4=\lambda^{2}+\lambda-2=0
$$

The eigenvalues are the roots of this equation, $\lambda_{1}=1$ and $\lambda_{2}=-2$. This means that expressions of the form $C e^{t}$ and $C e^{-2 t}$ (for $x$ and for $y$) give particular solutions to the system. Let's find any eigenvectors of matrix $A$ corresponding to these numbers.

An eigenvector of matrix $A$ corresponding to the eigenvalue $\lambda$ is any non-trivial solution to the homogeneous system $(A-\lambda E) U=0$, where $E$ is the identity matrix of the same dimension as matrix $A$, and $U$ is a column vector of unknowns. In our case, $A-\lambda E=\left(\begin{array}{cc}2-\lambda & -4 \\ 1 & -3-\lambda\end{array}\right), U=\binom{u_{1}}{u_{2}}$.

1) For $\lambda_{1}=1$, the homogeneous system $\left(\begin{array}{ll}1 & -4 \\ 1 & -4\end{array}\right)\binom{u_{1}}{u_{2}}=\binom{0}{0}$ reduces to one equation $u_{1}-4 u_{2}=0$, i.e., $u_{1}=4 u_{2}$. Let's set, for example, $u_{2}=1$; then $u_{1}=4$ and $U=\binom{4}{1}$ is an eigenvector corresponding to $\lambda=1$. This means that $\left(x_{1}=4 e^{t}, y_{1}=e^{t}\right)$ is a particular solution to the system.
2) For the eigenvalue $\lambda_{2}=-2$, the homogeneous system $\left(\begin{array}{ll}4 & -4 \\ 1 & -1\end{array}\right)\binom{u_{1}}{u_{2}}=\binom{0}{0}$, or the equation $u_{1}-u_{2}=0$; setting $u_{2}=1$, we get $u_{1}=1$ and the eigenvector $U=\binom{1}{1}$. This means that $\left(x_{2}=e^{-2 t}, y_{2}=e^{-2 t}\right)$ is another particular solution to the system.

The general solution to the system is a linear combination of its particular solutions with arbitrary constants (denoted here as $B_{1}$ and $B_{2}$): $x=B_{1} x_{1}+B_{2} x_{2}, y=B_{1} y_{1}+B_{2} y_{2}$.

Answer. $\left\{\begin{array}{l}x=4 B_{1} e^{t}+B_{2} e^{-2 t}, \\ y=B_{1} e^{t}+B_{2} e^{-2 t} .\end{array}\right.$

Note. Notice that the general solution obtained here looks different from the solution obtained earlier. In fact, they coincide if we take $C_{1}=4 B_{1}, C_{2}=B_{2}$ (or $B_{1}=\frac{C_{1}}{4}, B_{2}=C_{2}$).