Example. Solve the Cauchy problem

\[
\begin{cases}x^{\prime}=x+2 y-9 t, & x(0)=1 \\ y^{\prime}=2 x+y+4 e^{t}, & y(0)=2\end{cases}
\]
Solution. Assuming that the functions $x(t)$ and $y(t)$ are originals, denote their images as $X(p)$ and $Y(p)$.

1. By the theorem on differentiation of originals

$x^{\prime}(t) \longleftrightarrow p X(p)-x(0)=p X(p)-1, \quad y^{\prime}(t) \longleftrightarrow p Y(p)-y(0)=p Y(p)-2$.

By the linearity property, we find the images of the right-hand sides of the system of equations:

$$
\begin{gathered}
x+2 y-9 t \longleftrightarrow X(p)+2 Y(p)-\frac{9}{p^{2}} \\
2 x+y+4 e^{t} \longleftrightarrow 2 X(p)+Y(p)+\frac{4}{p-1}
\end{gathered}
$$

2. We form a system of operator equations:

$$
\left\{\begin{array}{l}
p X(p)-1=X(p)+2 Y(p)-\frac{9}{p^{2}} \\
p Y(p)-2=2 X(p)+Y(p)+\frac{4}{p-1}
\end{array}\right.
$$

3. We solve the system of operator equations:

$$
\begin{aligned}
& X(p)=\frac{4}{p-1}-\frac{3}{p}-\frac{3}{p^{2}}-\frac{8}{p(p-1)}+\frac{8}{(p-1)(p+1)}+\frac{4}{(p-3)(p-1)} \\
& Y(p)=-\frac{4}{p}+\frac{6}{p^{2}}+\frac{4}{p+1}+\frac{2}{p-3}
\end{aligned}
$$

4. We restore the originals from the images $X(p)$ and $Y(p)$:

$$
\begin{aligned}
& x(t)=2 e^{3 t}-4 e^{-t}-2 e^{t}+5-3 t \\
& y(t)=-4+6 t+4 e^{-t}+2 e^{3 t}
\end{aligned}
$$

5. By substituting $x(t)$ and $y(t)$ into the original system, both equations become identities. By calculating $x(0)$ and $y(0)$, we verify that $x(t)$ and $y(t)$ satisfy the initial conditions.

Answer.

$$
x(t)=2 e^{3 t}-4 e^{-t}-2 e^{t}+5-3 t, \quad y(t)=-4+6 t+4 e^{-t}+2 e^{3 t} .
$$

Conditions of the problems. Solve the Cauchy problems for systems of differential equations.

1. $\begin{cases}x^{\prime}=y-1, & x(0)=1, \\ y^{\prime}=-x-2 y, & y(0)=-1 .\end{cases}$
2. $\begin{cases}x^{\prime}=-y+2, & x(0)=-1, \\ y^{\prime}=x+1, & y(0)=0 .\end{cases}$
3. $\begin{cases}3 x^{\prime}+2 x+y^{\prime}=1, & x(0)=0, \\ x^{\prime}+4 y^{\prime}+3 y=0, & y(0)=0 .\end{cases}$
4. $\begin{cases}x^{\prime}-2 y=0, & x(0)=2, \\ y^{\prime}-2 x=0, & y(0)=2 .\end{cases}$
5. $\begin{cases}x^{\prime}=3 x+4 y, & x(0)=1, \\ y^{\prime}=4 x-3 y, & y(0)=1 .\end{cases}$
6. $\begin{cases}x^{\prime}=2 x+3 y+1, & x(0)=-1, \\ y^{\prime}=4 x-2 y, & y(0)=0 .\end{cases}$
7. $\begin{cases}x^{\prime}=-x+3 y+1, & x(0)=1, \\ y^{\prime}=x+y, & y(0)=2 .\end{cases}$
8. $\begin{cases}x^{\prime}+y=0, & x(0)=1, \\ y^{\prime}+x=0, & y(0)=-1 .\end{cases}$
9. $\left\{\begin{array}{ll}x^{\prime}=-y, & x(0)=1, \\ y^{\prime}=2 x+2 y, & y(0)=1 .\end{array} \quad 10 . \begin{cases}x^{\prime}+y^{\prime}-y=e^{t}, & x(0)=0, \\ 2 x^{\prime}+y^{\prime}+2 y=\cos t, & y(0)=0 .\end{cases}\right.$

## Answers.

1. $x(t)=3 e^{-t}+t e^{-t}-2, y(t)=1-2 e^{-t}-t e^{-t}$.
2. $x(t)=2 \sin t-1, y(t)=2-2 \cos t$.
3. $x(t)=\frac{1}{2}-\frac{1}{5} e^{-t}-\frac{3}{10} e^{-6 t / 11}, y(t)=\frac{1}{5}\left(e^{-t}-e^{-6 t / 11}\right)$.
4. $x(t)=\frac{5}{2} e^{2 t}-\frac{1}{2} e^{-2 t}, y(t)=\frac{5}{2} e^{2 t}-\frac{1}{2} e^{-2 t}$.
5. $x(t)=\frac{6}{5} e^{5 t}-\frac{1}{5} e^{-5 t}, y(t)=\frac{3}{5} e^{5 t}+\frac{2}{5} e^{-5 t}$.
6. $x(t)=-\frac{1}{8}-\frac{9}{16} e^{4 t}-\frac{5}{16} e^{-4 t}, y(t)=-\frac{1}{4}-\frac{3}{8} e^{4 t}+\frac{5}{8} e^{-4 t}$.
7. $x(t)=\frac{15}{8} e^{2 t}-\frac{9}{8} e^{-2 t}+\frac{1}{4}, y(t)=\frac{15}{8} e^{2 t}+\frac{3}{8} e^{-2 t}-\frac{1}{4}$.
8. $x(t)=e^{t}, y(t)=-e^{t}$.
9. $x(t)=e^{t}(\cos t+4 \sin t), y(t)=e^{t}(\cos t+3 \sin t)$.
10. $x(t)=e^{t}-\frac{11}{34} e^{4 t}-\frac{3}{17} \cos t+\frac{5}{17} \sin t-\frac{1}{2}$,

$$
y(t)=-\frac{2}{3} e^{t}+\frac{22}{51} e^{4 t}+\frac{4}{17} \cos t-\frac{1}{17} \sin t
$$

## $\Gamma$ lav a 3

## FOURIER SERIES

When studying the topic of FOURIER SERIES, you will learn to expand functions into trigonometric Fourier series and into Fourier series by arbitrary orthogonal systems on various finite and infinite intervals.

The main difficulty in studying the topic of FOURIER SERIES is the calculation of integrals. The package RESHEBNIK.VM will help you calculate integrals and thus significantly facilitate the study of the topic. Later, for example, when solving equations of mathematical physics, you will be able to obtain Fourier series expansions of the functions of interest by pressing a few buttons on the computer keyboard, using the capabilities of the STEM Plus module included in the RESHEBNIK.VM package.

## 3.1. Trigonometric Fourier series of the function $f(x)$ on the interval $(-\pi, \pi)$

Problem statement. Expand the function $y=f(x)$ into a trigonometric Fourier series on the interval $(-\pi, \pi)$.

PLAN OF SOLUTION. The trigonometric Fourier series on the interval $(-\pi, \pi)$ is the series

$$
a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n x+b_{n} \sin n x\right)
$$

The functions $1, \cos x, \cos 2 x, \ldots, \cos n x, \ldots, \sin x, \sin 2 x, \ldots, \sin n x, \ldots$ form an orthogonal basis in the space of functions defined on the interval $(-\pi, \pi)$, with the scalar product

$$
(u, v)=\int_{-\pi}^{\pi} u(x) v(x) d x
$$

The function $f$ belongs to this space if and only if

$$
(f, f)=\int_{-\pi}^{\pi} f(x)^{2} d x < \infty
$$

In this case, using the formulas for reducing the degree, the problem of expanding into a trigonometric Fourier series for functions of the form $f(x) = P(\sin x, \cos x)$, where $P(u, v)$ is a polynomial, is reduced.