4.205. Investigate the stability by the first approximation

$$
\frac{d x}{d t}=2-e^{y}-3 x-\cos x, \frac{d y}{d t}=2 y+8 \sin x
$$
The problem is solved. Let's write down the first three terms of the Maclaurin series expansion for the exponential, sine, and cosine functions.

$$
\begin{gathered}
e^{y}=1+y+\frac{y^{2}}{2}+\ldots ; \cos y=1-\frac{y^{2}}{2}+\frac{y^{4}}{24}-\ldots \\
\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\ldots
\end{gathered}
$$

If we limit ourselves to terms of only the first order of smallness and substitute these functions with \( e^{y} \approx 1+y \); \( \cos x \approx 1 \); \( \sin x \approx x \), then the system can be rewritten as:

$$
\frac{d x}{d t}=-3 x-y, \quad \frac{d y}{d t}=8 x+2 y
$$

We form the characteristic equation:

$$
\left|\begin{array}{cc}
(-3-\hat{\lambda}) & -1 \\
8 & (2-\hat{\lambda})
\end{array}\right|=\lambda^{2}+\lambda+2=0
$$

The roots of the characteristic equation are complex conjugates, the real part is negative, and the imaginary part is non-zero:

$$
\left.\lambda_{1,2}=-\frac{1}{2} \pm \frac{\sqrt{7} i}{2}, \alpha=-\frac{1}{2}<0, \quad \beta=\frac{\sqrt{7}}{2} \neq 0 \text { (case } 2 a\right)
$$

The equilibrium point is stable.