Problem 18
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 $$\frac{dx}{dt}=x-y,\qquad\frac{dy}{dt}=2x+3y.$$

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

The system is linear and homogeneous with coefficient matrix $$A=\begin{pmatrix}1&-1\\2&3\end{pmatrix}.$$

Step 2 · Characteristic equation and eigenvalues

$$\det(A-\lambda I)=0\implies(1-\lambda)(3-\lambda)+2=0 \implies\lambda^2-4\lambda+5=0.$$ Discriminant $\Delta=-4<0$, so $$\lambda_{1,2}=2\pm i.$$

Step 3 · Classification of the rest point

The eigenvalues are complex with positive real part: $\operatorname{Re}(\lambda_{1,2})=2>0$. The rest point $(0,0)$ is an unstable focus. All trajectories spiral away from the origin as $t\to+\infty$.

Step 4 · Direction of rotation

Evaluating $\dot y$ at $(1,0)$ gives $\dot y=2>0$, so the trajectories rotate counter-clockwise.

Step 5 · Eigenvector and general real solution

For $\lambda_1=2+i$: eigenvector $\mathbf v^{(1)}=(1,-1-i)^\top =(1,-1)^\top+i(0,-1)^\top$. With $\alpha=2$, $\beta=1$, $\mathbf p=(1,-1)^\top$, $\mathbf q=(0,-1)^\top$: $$x(t)=e^{2t}(C_1\cos t+C_2\sin t),$$ $$y(t)=e^{2t}\bigl[(-C_1+C_2)\cos t+(-C_1-C_2)\sin t\bigr].$$

Step 6 · Behaviour of trajectories

The distance from the origin satisfies $r(t)\propto e^{2t}\to+\infty$ as $t\to+\infty$, confirming outward spiraling. As $t\to-\infty$, $r(t)\to0$ and trajectories approach the origin while completing infinitely many counter-clockwise revolutions with angular frequency $\omega=1$ (period $T=2\pi$).

Step 7 · Stability analysis

Since $\operatorname{Re}(\lambda_{1,2})=2>0$, the rest point is unstable: it is not Lyapunov stable, since trajectories starting arbitrarily close to the origin eventually leave any prescribed neighbourhood.

Step 8 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{unstable focus}.}\quad \lambda_{1,2}=2\pm i,\quad \operatorname{Re}(\lambda)=2>0. }$$ All trajectories spiral counter-clockwise away from the origin with angular frequency $\omega=1$ and exponential growth rate $\alpha=2$.


Summary Table

QuantityValue
Eigenvalues$\det(A-\lambda I)=0\implies(1-\lambda)(3-\lambda)+2=0 \implies\lambda^2-4\lambda+5=0.$
Classificationunstable focus

Additional Notes

Result

The final classification is:

$$\boxed{ (0,0)\text{ is an \textbf{unstable focus}.}\quad \lambda_{1,2}=2\pm i,\quad \operatorname{Re}(\lambda)=2>0. }$$