Problem 20
Easy Equilibrium Points

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

QuantityValue
Eigenvalues$\det(A-\lambda I)=0\implies(-2-\lambda)(1-\lambda)+3=0 \implies\lambda^2+\lambda+1=0.$
Classificationstable focus

Additional Notes

Result

The final classification is:

$$\boxed{ (0,0)\text{ is a \textbf{stable focus} (asymptotically stable).}\quad \lambda_{1,2}=-\tfrac{1}{2}\pm\tfrac{\sqrt{3}}{2}\,i. }$$