Example 7. Solve the system of equations

$$
\frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x
$$

with initial conditions

$$
x(0)=x_{0}, \quad y(0)=y_{0}
$$
Solution. Differentiating the first equation of system (10) once with respect to $t$ and substituting $\frac{d y}{d t}=-x$ into the resulting equation, we reduce system (10) to a single second-order equation $\frac{d^{2} x}{d t^{2}}+x=0$, the general solution of which is

$$
x=C_{1} \cos t+C_{2} \sin t
$$

Since $y=\frac{d x}{d t}$, then $y=-C_{1} \sin t+C_{2} \cos t$; thus, the general solution of system (10) is:

$$
x=C_{1} \cos t+C_{2} \sin t, \quad y=-C_{1} \sin t+C_{2} \cos t
$$

A particular solution of system (10) that satisfies the initial conditions (11) is

$$
x=x_{0} \cos t+y_{0} \sin t, \quad y=-x_{0} \sin t+y_{0} \cos t
$$

By eliminating $t$ from equations (13) (by squaring and adding term by term), we obtain the phase trajectory

$$
x^{2}+y^{2}=R^{2}
$$

where $R=\sqrt{x_{0}^{2}+y_{0}^{2}}$. This is a circle passing through the point $M_{0}\left(x_{0}, y_{0}\right)$. Representing equations (13) in the form

$$
\begin{aligned}
& x=R \sin (t+\alpha) \\
& y=R \cos (t+\alpha)
\end{aligned}
$$

where $R=\sqrt{x_{0}^{2}+y_{0}^{2}}$, $\sin \alpha=\frac{x_{0}}{R}$, $\cos \alpha=\frac{y_{0}}{R}$, we notice that equations (15) express the time dependence of the current coordinates of the point $M(x(t), y(t))$, or simply $M(t)$, which starts its motion at $t=0$ from the point $M_{0}\left(x_{0}, y_{0}\right)$ and moves along the circle (14) (Fig. 29, a).

The direction of motion of the point $M(t)$ will be determined using the given system (10). For $x>0$, according to the equation $\frac{d y}{d t}=-x$, the value of $y$ decreases (as, for example, at the point $M_{1}(t)$), and for $x<0$ the value of $y$ increases (as, for example, at the point $M_{2}(t)$). Thus, the point $M(t)$ moves along the curve (14) clockwise. By arbitrarily changing the initial conditions (11) (while remaining within physically permissible limits), i.e., by arbitrarily changing the position of the initial point $M_{0}\left(x_{0}, y_{0}\right)$, we obtain all possible phase trajectories (14).

Now let's give another interpretation of equations (15) (or, equivalently, equations (13)). In three-dimensional space, take a right-handed Cartesian coordinate system $O x y z$. It is easy to verify that the point $N(x(t), y(t), z(t))$ (or simply $N(t)$) with coordinates (Fig. 29, b)

$$
\begin{gathered}
x(t)=R \sin (t+\alpha) \\
y(t)=R \cos (t+\alpha), \quad z(t)=t
\end{gathered}
$$

begins its motion at $t=0$ from the initial point $N_{0}\left(x_{0}, y_{0}, 0\right)$ and, as $t$ increases, moves along a helical line (16) located on a cylinder with the circle (14) as its base and generators parallel to the $O z$ axis.

Obviously, the point $N_{0}$ coincides with the point $M_{0}$ and that at any $t$ the point $N(t)$ projects onto the point $M(t)$ on the phase trajectory. Since the point $M(t)$ moves clockwise, the integral curve described by the point $N(t)$ is a left-handed helical line on the cylinder with the circle (14) as its base and generators parallel to the $O z$ axis. For different positions of the point $N_{0}\left(x_{0}, y_{0}, 0\right)$, the integral curves of system (10) corresponding to different values of $R=\sqrt{x_{0}^{2}+y_{0}^{2}}$ project onto the $x O y$ plane as different curves (14), and the integral curves corresponding to the same value of $R$ project onto the same curve (14).

An integral of a normal system (3) is a function $\psi\left(t, x_{1}\right.$, $x_{2}, \ldots, x_{n}$ ), defined and continuous along with its first-order partial derivatives $\frac{\partial \psi}{\partial t}, \frac{\partial \psi}{\partial x_{k}}, k=1,2, \ldots, n$, in the domain $D$, if, when substituted with any solution $x_{1}(t), x_{2}(t), \ldots, x_{n}(t)$ of system (3), it takes a constant value, i.e., the function $\psi\left(t, x_{1}\right.$, $\left.x_{2}, \ldots, x_{n}\right)$ depends only on the choice of the solution $x_{1}(t), x_{2}(t), \ldots, x_{n}(t)$, but not on the variable $t$.

## A first integral of a normal system (3) is an equality

$$
\psi\left(t, x_{1}, x_{2}, \ldots, x_{n}\right)=C
$$

where $\psi\left(t, x_{1}, x_{2}, \ldots, x_{n}\right)$ is an integral of system (3), and $C$ is an arbitrary constant.