Problem 23
Easy Linearization

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

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

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

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

Step 2 · Characteristic equation and eigenvalues

$$\det(A-\lambda I)=0\implies\lambda^2+2\lambda-5=0.$$ Discriminant $\Delta=24>0$, so $$\lambda_1=-1+\sqrt{6}\approx1.449,\qquad\lambda_2=-1-\sqrt{6}\approx-3.449.$$

Step 3 · Classification of the rest point

The eigenvalues are real, distinct, and of opposite sign. The rest point $(0,0)$ is a saddle point: it is unstable.

Step 4 · Eigenvectors and separatrices

For $\lambda_1=-1+\sqrt6$: eigenvector $(\sqrt6-2,1)^\top$, giving the unstable separatrix tangent to $y=\tfrac{\sqrt6+2}{2}x$ (departs the origin as $t\to+\infty$). For $\lambda_2=-1-\sqrt6$: eigenvector $(-(2+\sqrt6),1)^\top$, giving the stable separatrix (attracted to the origin as $t\to+\infty$).

Step 5 · General solution

$$x(t)=C_1(\sqrt6-2)e^{\lambda_1 t}-C_2(2+\sqrt6)e^{\lambda_2 t},\qquad y(t)=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t}.$$

Step 6 · Behaviour of integral curves

Eliminating $t$ gives the homogeneous ODE $dy/dx=(x+y)/(-3x+2y)$, whose integral curves form the characteristic hyperbolic family of a saddle, asymptotic to the two eigendirections found above.

Step 7 · Stability analysis

Since $\lambda_1=-1+\sqrt6>0$, the rest point is unstable: there exist trajectories (all those not on the stable separatrix) that diverge from the origin.

Step 8 · Conclusion

$$\boxed{ (0,0)\text{ is a \textbf{saddle point} (unstable).} }$$ The eigenvalues $\lambda_1=-1+\sqrt6\approx1.449>0$ and $\lambda_2=-1-\sqrt6\approx-3.449<0$ are real with opposite signs. The unstable separatrix is tangent to $\mathbf v^{(1)}=(\sqrt6-2,1)^\top$ and the stable separatrix to $\mathbf v^{(2)}=(-(2+\sqrt6),1)^\top$ at the origin. All other trajectories exhibit the characteristic hyperbolic divergence of a saddle.


Summary Table

QuantityValue
Eigenvalues$\det(A-\lambda I)=0\implies\lambda^2+2\lambda-5=0.$
Classificationsaddle point

Additional Notes

Result

The final classification is:

$$\boxed{ (0,0)\text{ is a \textbf{saddle point} (unstable).} }$$