Example 6. Find the solution to the system

$$
\left\{\begin{array}{l}
\frac{d x}{d t}=-7 x+y+5 \\
\frac{d y}{d t}=-2 x-5 y-37 t
\end{array}\right.
$$

satisfying the initial condition $x(0)=0, y(0)=0$.
Solution. Since $5 \risingdotseq \frac{5}{p}, -37 t = -\frac{37}{p^{2}}$ and $x_{0}=y_{0}=0$, the operator system will have the form

$$
\left\{\begin{array}{l}
p X(p) = -7 X(p) + Y(p) + \frac{5}{p} \\
p Y(p) = -2 X(p) - 5 Y(p) + \frac{37}{p^{2}}
\end{array}\right.
$$

Solving this, we get

$$
X(p) = \frac{5 p^{2} + 25 p - 37}{p^{2} \left(p^{2} + 12 p + 37\right)}, \quad Y(p) = \frac{-47 p - 259}{p^{2} \left(p^{2} + 12 p + 37\right)}
$$

Decomposing the fractions on the right-hand sides into elementary fractions:

$$
\begin{aligned}
& X(p) = \frac{1}{p} - \frac{1}{p^{2}} - \frac{p + 6}{p^{2} + 12 p + 37} \\
& Y(p) = \frac{1}{p} - \frac{7}{p^{2}} - \frac{p + 5}{p^{2} + 12 p + 37}
\end{aligned}
$$

or

$$
\begin{aligned}
& X(p) = \frac{1}{p} - \frac{1}{p^{2}} - \frac{p + 6}{(p + 6)^{2} + 1} \\
& Y(p) = \frac{1}{p} - \frac{7}{p^{2}} - \frac{p + 6}{(p + 6)^{2} + 1} + \frac{1}{(p + 6)^{2} + 1}
\end{aligned}
$$

Transitioning to the originals, we get the desired solution

$$
x(t) = 1 - t - e^{-6 t} \cos t, \quad y(t) = 1 - 7 t + e^{-6 t} \cos t + e^{-6 t} \sin t.
$$

In the following problems, solve the systems of equations using the operational method:

866. $\begin{cases}\frac{d x}{d t} + y = 0, & x(0) = 2, \\ \frac{d y}{d t} + x = 0, & y(0) = 0 .\end{cases}$
867. $\begin{cases}\frac{d x}{d t} + x - 2 y = 0, & x(0) = 1, \\ \frac{d y}{d t} + x + 4 y = 0, & y(0) = 1 .\end{cases}$
868. $\left\{\begin{array}{l}\frac{d x}{d t} = -y, \quad x(0) = 1, \\ \frac{d y}{d t} = 2(x + y), \quad y(0) = 1 .\end{array}\right.$
869. $\begin{cases}\frac{d x}{d t} + 2 y = 3 t, & x(0) = 2, \\ \frac{d y}{d t} - 2 x = 4, & y(0) = 3 .\end{cases}$
870. $\begin{cases}\frac{d x}{d t} + x = y + e^{t}, & x(0) = 1, \\ \frac{d y}{d t} + y = x + e^{t}, & y(0) = 1 .\end{cases}$
871. $\left\{\begin{array}{l}\frac{d x}{d t} - \frac{d y}{d t} = -\sin t, \quad x(0) = \frac{1}{2}, \\ \frac{d x}{d t} + \frac{d y}{d t} = \cos t, \quad y(0) = -\frac{1}{2} .\end{array}\right.$
872. $\begin{cases}\frac{d x}{d t} = y - z, & x(0) = 1, \\ \frac{d y}{d t} = x + y, & y(0) = 2, \\ \frac{d z}{d t} = x + z, & z(0) = 3 .\end{cases}$
873. $\left\{\begin{array}{l}\frac{d x}{d t} = 4 y + z, \quad x(0) = 5, \\ \frac{d y}{d t} = z, \quad y(0) = 0, \\ \frac{d z}{d t} = 4 y, \quad z(0) = 4 .\end{array}\right.$
874. $\left\{\begin{array}{l}\frac{d x}{d t} + 2 \frac{d y}{d t} + x + y + z = 0, \quad x(0) = 1, \\ \frac{d x}{d t} + \frac{d y}{d t} + x + z = 0, \quad y(0) = 1, \\ \frac{d z}{d t} - 2 \frac{d y}{d t} - y = 0, \quad z(0) = -2 .\end{array}\right.$
875. $\left\{\begin{array}{l}\frac{d x}{d t} - \frac{d y}{d t} - 2 x + 2 y = 1 - 2 t, \quad x(0) = 0, \quad y(0) = 0, \\ \left.\frac{d^{2} x}{d 2} + 2 y y + x\right)\end{array}\right.$

$$
\frac{d^{2} x}{d t^{2}} + 2 \frac{d y}{d t} + x = 0, \quad x^{\prime}(0) = 0
$$

876. $\begin{cases}\frac{d^{2} x}{d t^{2}} = y, & x(0) = y(0) = 1, \\ \frac{d^{2} y}{d t^{2}} = x, & x^{\prime}(0) = 2, \quad y^{\prime}(0) = 0 .\end{cases}$
877. $\begin{cases}\frac{d^{2} x}{d t^{2}} = x - 4 y, & x(0) = 2, \quad y(0) = 0, \\ \frac{d^{2} y}{d t^{2}} = -x + y, & x^{\prime}(0) = -\sqrt{3}, \quad y^{\prime}(0) = \frac{\sqrt{3}}{2} \text {. }\end{cases}$
878. 

$$
\left\{\begin{array}{l}
\frac{d^{2} x}{d t^{2}} + \frac{d y}{d t} = e^{t} - x, \quad x(0) = 1, \quad y(0) = 0 \\
\frac{d^{2} y}{d t} + \frac{d x}{d t} = 1, \quad x^{\prime}(0) = 2, \quad y^{\prime}(0) = -1
\end{array}\right.
$$

879. $\left\{\begin{array}{l}\frac{d^{2} x}{d t^{2}} + x + y = 5, \quad x(0) = x^{\prime}(0) = 0, \\ \frac{d^{2} y}{d t^{2}} - 4 x - 3 y = -3, \quad y(0) = y^{\prime}(0) = 0 .\end{array}\right.$
880. $\left\{\begin{array}{l}\frac{d x}{d t} + 4 y + 2 x = 4 t + 1, \quad x(0) = 0, \\ \frac{d y}{d t} + x - y = \frac{3}{2} t^{2}, \quad y(0) = 0 .\end{array}\right.$
881. $\left\{\begin{array}{l}\frac{d x}{d t} + y - 2 x = 0, \quad x(0) = 2, \\ \frac{d y}{d t} + x - 2 y = -5 e^{t} \sin t, \quad y(0) = 3 .\end{array}\right.$

## Theory

4

## Stability

## § 25. Lyapunov Stability. Basic Concepts and Definitions

Consider the system of differential equations

$$
\frac{d x_{i}}{d t} = f_{i}(x_{1}, x_{2}, \ldots, x_{n}, t), \quad i = 1, 2, \ldots, n
$$

A solution $\varphi_{i}(t), i = 1, 2, \ldots, n$, of system (1), satisfying the initial conditions $\varphi_{i}(t_{0}) = \varphi_{i 0}, i = 1, 2, \ldots, n$, is called stable in the sense of Lyapunov as $t \rightarrow \infty$ if for any $\varepsilon > 0$ there exists $\delta(\vare