Examine the character of the rest point (0,0) of the system…

Examine the character of the rest point $(0,0)$ of the system $$\dot{x}=-x+2y,\qquad\dot{y}=-2x+3y.$$

Step 1 · Coefficient matrix

$$A=\begin{pmatrix}-1&2\\-2&3\end{pmatrix}.$$

Step 2 · Characteristic equation and eigenvalues

$$\det(A-\lambda I)=0\implies(-1-\lambda)(3-\lambda)+4=0 \implies\lambda^2-2\lambda+1=0\implies(\lambda-1)^2=0.$$ There is a repeated eigenvalue $\lambda_1=\lambda_2=1$.

Step 3 · Eigenspace

Solving $(A-I)\mathbf v=0$ gives $v_1=v_2$, so there is only one independent eigenvector $\mathbf v^{(1)}=(1,1)^\top$: the geometric multiplicity (1) is less than the algebraic multiplicity (2), so $A$ is defective and a generalised eigenvector is required.

Step 4 · Generalised eigenvector

Solving $(A-I)\mathbf w=\mathbf v^{(1)}$ gives $\mathbf w=(0,\tfrac12)^\top$.

Step 5 · General solution

Using the Jordan chain $\{\mathbf v^{(1)},\mathbf w\}$: $$x(t)=e^{t}(C_1+C_2\,t),\qquad y(t)=e^{t}\!\left(C_1+\tfrac{1}{2}C_2+C_2\,t\right).$$

Step 6 · Phase portrait analysis

Since $y(t)-x(t)=\tfrac12C_2e^t$ while $x(t)=e^t(C_1+C_2t)$, for $C_2\neq0$ the ratio $(y-x)/x\to0$ as $t\to+\infty$, so all non-trivial trajectories approach the line $y=x$ asymptotically (in direction). This is the hallmark of an improper (degenerate) unstable node.

Step 7 · Behaviour of trajectories

Since $\lambda=1>0$: as $t\to+\infty$, $|x(t)|,|y(t)|\to+\infty$, so all trajectories diverge. As $t\to-\infty$, $e^t\to0$ dominates the polynomial factor, so trajectories approach the origin tangentially to $y=x$. The unique straight-line trajectory ($C_2=0$) lies exactly on $y=x$.

Step 8 · Stability analysis

Since $\lambda=1>0$, the rest point is unstable: it is not Lyapunov stable.

Step 9 · Conclusion

$$\boxed{ (0,0)\text{ is an \textbf{unstable improper (degenerate) node}.}\quad \lambda_{1,2}=1\;(\text{repeated}). }$$ The repeated positive eigenvalue $\lambda=1$ with a single eigenvector $\mathbf{v}^{(1)}=(1,1)^{\top}$ produces an improper node. All trajectories approach the origin tangentially to the line $y=x$ as $t\to-\infty$ and diverge to infinity as $t\to+\infty$, confirming instability.

Summary Table

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.