Example 2. Solve the system of equations $\left\{\begin{array}{l}x^{\prime}=-4 x-14 y, \\ y^{\prime}=5 x+13 y .\end{array}\right.$
Solution. Immediately note that the unknown functions are denoted by different letters, and we will consider the variable $t$ as the argument.

The essence of the method is to transition from the given system to another, equivalent system, in which differential groups or complete differentials can be separated. For example, if we perform the specified operations (multiplications by corresponding numbers and subsequent addition of the results) on the equations of the given system, we will sequentially obtain:

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-104.jpg?height=111&width=431&top_left_y=575&top_left_x=1375)

$$
\begin{aligned}
& \Rightarrow 5 x^{\prime}+7 y^{\prime}=15 x+21 y \Leftrightarrow \frac{d}{d t}(5 x+7 y)=3(5 x+7 y)
\end{aligned}
$$

This was the goal of our transformation. We will transform the last equality and integrate: $\int \frac{d(5 x+7 y)}{5 x+7 y}=3 \int d t$, $\ln |5 x+7 y|=3 t+\ln C_{1}, 5 x+7 y=C_{1} e^{3 t}$. We have obtained one relationship between the unknown functions, which we will rewrite as: $7 y=C_{1} e^{3 t}-5 x$.

We will form a new system by appending the last equality to the first equation of the original system:

$$
\left\{\begin{array}{l|l}
7 y=C_{1} e^{3 t}-5 x & 2 \\
-14 y=x^{\prime}+4 x . & 1
\end{array}+\right.
$$

By adding (as indicated in the diagram) these equations, we obtain a linear (non-homogeneous) first-order equation: $x^{\prime}-6 x=-2 C_{1} e^{3 t}$, which can be solved using the substitution method: $x=u v, x^{\prime}=u^{\prime} v+u v^{\prime}$. We obtain sequentially

$$
u^{\prime} v+u v^{\prime}-6 u v=-2 C_{1} e^{3 t} \Rightarrow\left\{\begin{array} { r l } 
{ v ^ { \prime } } & { = 6 v } \\
{ u ^ { \prime } v } & { = - 2 C _ { 1 } e ^ { 3 t } }
\end{array} \Rightarrow \left\{\begin{array}{rl}
v & =e^{6 t} \\
u & =\frac{2}{3} C_{1} e^{-3 t}+C_{2}
\end{array}\right.\right.
$$

$x=u v=\frac{2}{3} C_{1} e^{3 t}+C_{2} e^{6 t}$ - this is the first unknown function of the system. The second unknown function $y=y(t)$ is found from the relationship $7 y=$ $=C_{1} e^{3 t}-5 x$.

$$
\text { Answer. }\left\{\begin{array}{l}
x=\frac{2}{3} C_{1} e^{3 t}+C_{2} e^{6 t} \\
y=-\frac{1}{3} C_{1} e^{-3 t}-\frac{5}{7} C_{2} e^{6 t}
\end{array}\right.
$$