Example 4. Solve the Cauchy problem for the system

$$
\left\{\begin{array}{l}
\frac{d x}{d t}=8 y \\
\frac{d y}{d t}=-2 z \\
\frac{d z}{d t}=2 x+8 y-2 z
\end{array}\right.
$$

with initial conditions $x(0)=-4, y(0)=0, z(0)=1$.
Solution. Characteristic equation

$$
\left|\begin{array}{rrc}
-r & 8 & 0 \\
0 & -r & -2 \\
2 & 8 & -2-r
\end{array}\right|=0, \text { or }(r+2)\left(r^{2}+16\right)=0 \text {. }
$$

The roots of equation (17): $r_{1}=-2, r_{2}=4 i, r_{3}=-4 i$. The real root $r_{1}=-2$ corresponds to the solution

$$
x_{1}=\lambda_{1} e^{-2 t}, \quad y=\mu_{1} e^{-2 t}, \quad z_{1}=\nu_{1} e^{-2 t} .
$$

Substituting (18) into system (16) and canceling by $e^{-2 t}$, we get

$$
-2 \lambda_{1}=8 \mu_{1}, \quad-2 \mu_{1}=-2 \nu_{1}, \quad-2 \nu_{1}=2 \lambda_{1}+8 \mu_{1}-2 \nu_{1}
$$

from which $\lambda_{1}=-4 \mu_{1}, \nu_{1}=\mu_{1}$. Let, for example, $\mu_{1}=1$, then $\lambda_{1}=-4, \nu_{1}=1$ and the particular solution (18):

$$
x_{1}=-4 e^{-2 t}, \quad y_{1}=e^{-2 t}, \quad z_{1}=e^{-2 t} .
$$

The complex root $r_{2}=4 i$ corresponds to the solution

$$
x_{2}=\lambda_{2} e^{4 i t}, \quad y_{2}=\mu_{2} e^{4 i t}, \quad z_{2}=\nu_{2} e^{4 i t}
$$

substituting which into (16) and canceling by $e^{4 \text { іit }}$, we get

$$
4 i \lambda_{2}=8 \mu_{2}, \quad 4 i \mu_{2}=-2 \nu_{2}, \quad 4 i \nu_{2}=2 \lambda_{2}+8 \mu_{2}-2 \nu_{2}
$$

from which $\lambda_{2}=-2 i \mu_{2}, \nu_{2}=-2 i \mu_{2}$, so, for example, at $\mu_{2}=i$ we have $\lambda_{2}=2$, $\nu_{2}=2$ and the particular solution

$$
x_{2}=2 e^{4 i t}, \quad y_{2}=i e^{4 i t}, \quad z_{2}=\nu_{2} e^{4 i t}
$$

The root $r_{3}=-4 i$ corresponds to the solution, complex conjugate to the solution (20), i.e.

$$
x_{3}=2 e^{-4 i t}, \quad y_{3}=-i e^{-4 i t}, \quad z_{3}=\nu_{2} e^{-4 i t} .
$$

Considering (19), (20), (21), we get the general solution

$$
\begin{aligned}
& x=-4 C_{1} e^{-2 t}+2 C_{2} e^{4 i t}+2 C_{3} e^{-4 i t} \\
& y=C_{1} e^{-2 t}+C_{2} i e^{4 i t}-C_{3} i e^{-4 i t} \\
& z=C_{1} e^{-2 t}+2 C_{2} e^{4 i t}+2 C_{3} e^{-4 i t}
\end{aligned}
$$

Finally, let's isolate the solution with initial conditions $x(0)=-4, y(0)=0$, $z(0)=1$. From (22) at $t=0$ we have

$$
\left\{\begin{aligned}
-4 & =-4 C_{1}+2 C_{2}+2 C_{3} \\
0 & =C_{1}+C_{2} i-C_{3} i \\
1 & =C_{1}+2 C_{2}+2 C_{3}
\end{aligned}\right.
$$

from which $C_{1}=1, C_{2}=\frac{i}{2}, C_{3}=-\frac{i}{2}$. Thus,

$$
\left\{\begin{array}{l}
x=-4 e^{-2 t}+i e^{4 i t}-i e^{-4 i t} \\
y=e^{-2 t}-\frac{1}{2} e^{i 4 t}-\frac{1}{2} e^{-i 4 t} \\
z=e^{-2 t}+i e^{i 4 t}-i e^{-i 4 t}
\end{array}\right.
$$

Using Euler's formulas $e^{ \pm i \alpha t}=\cos \alpha t \pm i \sin \alpha t$, we finally get

$$
x=-4 e^{-2 t}-2 \sin 4 t, \quad y=e^{-2 t}-\cos 4 t, \quad z=e^{-2 t}-2 \sin 4 t .
$$

## Problems for independent solving

Using the Euler method, find the general solution of the given systems and, where specified, isolate the solution that satisfies the given initial conditions.

804. $\left\{\begin{array}{l}\frac{d x}{d t}=8 y-x \\ \frac{d y}{d t}=x+y .\end{array}\right.$
805. $\left\{\begin{array}{l}\frac{d x}{d t}=x-y, \\ \frac{d y}{d t}=y-x .\end{array}\right.$
806. $\left\{\begin{array}{l}\frac{d x}{d t}=2 x+y, \\ \frac{d y}{d t}=x-3 y, \quad x(0)=y(0)=0 .\end{array}\right.$
807. $\left\{\begin{array}{l}\frac{d x}{d t}=x+y, \quad x(0)=0, \\ \frac{d y}{d t}=4 y-2 x, \quad y(0)=-1 .\end{array}\right.$
808. $\left\{\begin{array}{l}\frac{d x}{d t}=4 x-5 y, \quad x(0)=0, \\ \frac{d y}{d t}=x, \quad y(0)=1 .\end{array}\right.$
809. $\left\{\begin{array}{l}\frac{d x}{d t}=-x+y+z, \\ \frac{d y}{d t}=x-y+z, \\ \frac{d z}{d t}=x+y-z .\end{array}\right.$
810. $\left\{\begin{array}{l}\frac{d x}{d t}=2 x-y+z \\ \frac{d y}{d t}=x+2 y-z \\ \frac{d z}{d t}=x-y+2 z .\end{array}\right.$
811. $\left\{\begin{array}{l}\frac{d x}{d t}=2 x-y+z, \quad x(0)=0, \\ \frac{d y}{d t}=x+z, \quad y(0)=0, \\ \frac{d z}{d t}=y-2 z-3 x, \quad z(0)=1 .\end{array}\right.$

## § 23. Methods of integrating non-homogeneous linear systems with constant coefficients

Let us have a non-h