Examine the character of the rest point (0,0) of the system…
Examine the character of the rest point $(0,0)$ of the system $$\dot{x}=-2x-3y,\qquad\dot{y}=x+y.$$
Explanation of Method
Since the system is already linear and homogeneous, its coefficient matrix is written down directly. The eigenvalues are computed from the characteristic equation, and the rest point is classified according to the standard theory of planar linear systems: a node, saddle, focus, or centre, depending on the sign and nature of the eigenvalues.
Step 1 · Coefficient matrix
$$A=\begin{pmatrix}-2&-3\\1&1\end{pmatrix}.$$
Step 2 · Characteristic equation and eigenvalues
$$\det(A-\lambda I)=0\implies(-2-\lambda)(1-\lambda)+3=0 \implies\lambda^2+\lambda+1=0.$$ Discriminant $\Delta=-3<0$, so $$\lambda_{1,2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}\,i.$$
Step 3 · Classification of the rest point
The eigenvalues are complex with negative real part: $\operatorname{Re}(\lambda_{1,2})=-\tfrac12<0$. The rest point $(0,0)$ is a stable focus. All trajectories spiral toward the origin as $t\to+\infty$.
Step 4 · Direction of rotation
Evaluating $\dot y$ at $(1,0)$ gives $\dot y=1>0$, so the trajectories rotate counter-clockwise.
Step 5 · Eigenvector and general real solution
For $\lambda_1=-\tfrac12+\tfrac{\sqrt3}{2}i$: eigenvector $\mathbf v^{(1)}=(-1+\sqrt3 i,\,2)^\top=(-1,2)^\top+i(\sqrt3,0)^\top$. With $\alpha=-\tfrac12$, $\beta=\tfrac{\sqrt3}{2}$: $$x(t)=e^{-t/2}\!\left[(-C_1+\sqrt3C_2)\cos\tfrac{\sqrt3}{2}t +(-C_2-\sqrt3C_1)\sin\tfrac{\sqrt3}{2}t\right],$$ $$y(t)=e^{-t/2}\!\left[2C_1\cos\tfrac{\sqrt3}{2}t+2C_2\sin\tfrac{\sqrt3}{2}t\right].$$
Step 6 · Behaviour of trajectories
The distance from the origin satisfies $r(t)\propto e^{-t/2}\to0$ as $t\to+\infty$: all trajectories spiral inward. The angular frequency is $\omega=\tfrac{\sqrt3}{2}$, giving period $T=\tfrac{4\pi}{\sqrt3}$. As $t\to-\infty$, $r(t)\to+\infty$.
Step 7 · Stability analysis
Since $\operatorname{Re}(\lambda_{1,2})=-\tfrac12<0$, the rest point is both stable in the sense of Lyapunov and asymptotically stable: every trajectory converges to the origin as $t\to+\infty$.
Step 8 · Conclusion
$$\boxed{ (0,0)\text{ is a \textbf{stable focus} (asymptotically stable).}\quad \lambda_{1,2}=-\tfrac{1}{2}\pm\tfrac{\sqrt{3}}{2}\,i. }$$ All trajectories spiral counter-clockwise toward the origin with angular frequency $\omega=\tfrac{\sqrt{3}}{2}$ and exponential decay rate $\alpha=\tfrac{1}{2}$, confirming asymptotic stability.
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\det(A-\lambda I)=0\implies(-2-\lambda)(1-\lambda)+3=0 \implies\lambda^2+\lambda+1=0.$ |
| Classification | stable focus |
Additional Notes
- The classification follows from the standard theory of planar linear autonomous systems, based on the sign and nature of the eigenvalues of the coefficient matrix.
Result
The final classification is: