Example 1. Determine the character of the equilibrium point (0,0) of the system

\[
\left\{\begin{array}{l}
\frac{d x}{d t}=5 x-y \\
\frac{d y}{d t}=2 x+y
\end{array}\right.
\]
Solution. We form the characteristic equation

$$
\left|\begin{array}{cc}
5-\lambda & -1 \\
2 & 1-\lambda
\end{array}\right|=0, \quad \text { or } \quad \lambda^{2}-6 \lambda+7=0
$$

Its roots $\lambda_{1}=3+\sqrt{2}>0, \lambda_{2}=3-\sqrt{2}>0$ are real, distinct, and positive. Therefore, the equilibrium point (0,0) is an unstable node.

The relationship between the types of equilibrium points and the values of the roots of the characteristic equation (2) can be represented visually. For this, we introduce the notations $\sigma=-\left(a_{11}+a_{22}\right), \Delta=a_{11} a_{22}-a_{12} a_{21}$. Then the characteristic equation can be written as $\lambda^{2}+\sigma \lambda+\Delta=0$.

Consider the plane with rectangular Cartesian coordinates $\Delta$ and $\sigma$ and mark on it the regions corresponding to different types of equilibrium points (Fig. 42). From the classification given above, it follows that the conditions for the stability of the equilibrium point are $\operatorname{Re} \lambda_{1}0$ and $\sigma>0$, i.e., for points that are in the first quadrant.

If $\lambda_{1}$ and $\lambda_{2}$ are complex, then the equilibrium point will be of the focus type. This condition is satisfied by points that lie between the branches of the parabola $\sigma^{2}=4 \Delta$ and do not belong to the $O \Delta$ axis $\left(\sigma^{2}0$, correspond to equilibrium points of the center type.

Points located outside the parabola $\sigma^{2}=4 \Delta\left(\sigma^{2}>4 \Delta\right)$ correspond to equilibrium points of the node type.

The region of the $O \Delta \sigma$ plane where $\Delta<0$ contains equilibrium points of the saddle type.

Excluding special cases (passing through the origin), we note that a saddle can transition into a stable or unstable node (Fig. 42).

A stable node can transition either into a saddle or a stable focus. The case of equal roots $\lambda_{1}=\lambda_{2}$ corresponds to the boundary between nodes and foci, i.e., the parabola $\sigma^{2}=4 \Delta$.

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-202.jpg?height=508&width=743&top_left_y=153&top_left_x=321)

Fig. 42