Example 6. Given the general solution (6) of system (7), find the particular solution of this system that satisfies the initial conditions $x_{1}(0)=0$, $x_{2}(0)=-4$.
Solution. The problem reduces to finding such values of the constants $C_{1}$ and $C_{2}$ that the following relations hold:

$$
0=C_{1}+C_{2}, \quad-4=2 C_{1}-2 C_{2}
$$

Solving this system, we find $C_{1}=-1, C_{2}=1$. The desired particular solution is

$$
x_{1}(t)=-e^{-t}+e^{3 t}, \quad x_{2}(t)=-2 e^{-t}-2 e^{3 t}.
$$

Remark 1. Not every system of differential equations can be reduced to a single equation. For example, the system

$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=-x_{1} \\
\frac{d x_{2}}{d t}=x_{2}
\end{array}\right.
$$

splits into two independent equations. The general solution in this case is obtained by integrating each equation separately:

$$
x_{1}=C_{1} e^{-t}, \quad x_{2}=C_{2} e^{t}
$$

Remark 2. If the number of equations in the system is $n$, and the number of sought functions is $N$, with $N>n$, then such a system is indeterminate. In this case, one can arbitrarily choose $N-n$ sought functions (provided they are differentiable the required number of times) and find the remaining $n$ functions depending on them.

Remark 3. If the system consists of $n$ equations, and the number of sought functions is $N$, with $N<n$, then this system may be inconsistent, i.e., have no solution.

Consider, for simplicity, a normal system of two differential equations

$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=f_{1}\left(t, x_{1}, x_{2}\right) \\
\frac{d x_{2}}{d t}=f_{2}\left(t, x_{1}, x_{2}\right)
\end{array}\right.
$$

We will consider the system of values $t, x_{1}, x_{2}$ as rectangular Cartesian coordinates of a point in three-dimensional space, referred to a coordinate system $O t x_{1} x_{2}$. The solution

$$
x_{1}=x_{1}(t), \quad x_{2}=x_{2}(t)
$$

which takes the value $x_{1}^{0}, x_{2}^{0}$ at $t=t_{0}$, represents in this space a certain line passing through the point $M_{0}\left(t_{0}, x_{1}^{0}, x_{2}^{0}\right)$. This line is called an integral curve (line) of the normal system (9).

The Cauchy problem for the system (9) can be formulated geometrically as follows: in the space of variables $(t, x_{1}, x_{2})$, find an integral curve passing through the given point $\left(t_{0}, x_{1}^{0}, x_{2}^{0}\right)$. The Cauchy theorem establishes the existence and uniqueness of such a line.

The normal system (9) and its solution can be interpreted as follows. We will consider the independent variable $t$ as time, and the system of values of the functions $x_{1}, x_{2}$ as rectangular Cartesian coordinates of a point in the plane $x_{1} O x_{2}$. This plane of variables $x_{1} O x_{2}$ is called the phase plane. In the phase plane, the solution

$$
x_{1}=x_{1}(t), \quad x_{2}=x_{2}(t)
$$

of the system (9), which takes the initial values $x_{1}^{0}, x_{2}^{0}$ at $t=t_{0}$, is represented by a line $A B$

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-153.jpg?height=282&width=348&top_left_y=174&top_left_x=904)

Fig. 28 (Fig. 28), passing through the point $M_{0}\left(x_{1}^{0}, x_{2}^{0}\right)$. This line is called a trajectory of the system (phase trajectory). Clearly, the trajectory of the system (9) is the projection of the integral curve onto the phase plane.

The system (9) determines at each moment of time $t$ in a given point $x_{1}, x_{2}$ of the phase plane the coordinates of the velocity $\left\{f_{1}, f_{2}\right\}$ of the moving point.