Examine the character of the rest point (0,0) of the system…
Examine the character of the rest point $(0,0)$ of the system $$\dot{x}=3x-5y,\qquad\dot{y}=2x-3y.$$
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
$$A=\begin{pmatrix}3&-5\\2&-3\end{pmatrix}.$$
Step 2 · Characteristic equation and eigenvalues
$$\det(A-\lambda I)=0\implies(3-\lambda)(-3-\lambda)+10=0 \implies\lambda^2+1=0\implies\lambda_{1,2}=\pm i.$$
Step 3 · Classification of the rest point
The eigenvalues are purely imaginary. Since the system is linear (not merely linearised), the rest point is a centre exactly, not subject to the indeterminacy that arises for nonlinear systems.
Step 4 · Direction of rotation
Evaluating $\dot y$ at $(1,0)$ gives $\dot y=2>0$, so the trajectories rotate counter-clockwise.
Step 5 · General real solution
With eigenvector $\mathbf v=(3,2)^\top+i(1,0)^\top$ for $\lambda=i$: $$x(t)=(3C_1+C_2)\cos t+(-C_1+3C_2)\sin t,\qquad y(t)=2C_1\cos t+2C_2\sin t.$$
Step 6 · First integral and family of trajectories
Seeking a quadratic invariant $F=ax^2+bxy+cy^2$ with $\dot F\equiv0$ along trajectories gives (up to scale) $a=2$, $b=-6$, $c=5$, so the family of integral curves is $$2x^2-6xy+5y^2=K,\quad K>0,$$ a family of concentric ellipses centred at the origin (positive definite since the discriminant $36-40=-4<0$ with positive leading coefficient).
Step 7 · Behaviour of trajectories
Since $\operatorname{Re}(\lambda_{1,2})=0$, every trajectory is a closed ellipse; trajectories neither approach nor recede from the origin. The angular frequency is $\omega=1$, giving period $T=2\pi$.
Step 8 · Stability analysis
The rest point is stable in the sense of Lyapunov (trajectories remain on a fixed ellipse $2x^2-6xy+5y^2=K$ close to the origin for small $K$) but not asymptotically stable, since no trajectory converges to the origin.
Step 9 · Conclusion
$$\boxed{ (0,0)\text{ is a \textbf{centre} (stable, not asymptotically stable).}\quad \lambda_{1,2}=\pm i. }$$ All trajectories are closed ellipses $2x^2-6xy+5y^2=K$ ($K>0$) traversed counter-clockwise with period $T=2\pi$. The origin is Lyapunov stable but not asymptotically stable.
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\det(A-\lambda I)=0\implies(3-\lambda)(-3-\lambda)+10=0 \implies\lambda^2+1=0\implies\lambda_{1,2}=\pm i.$ |
| Classification | centre |
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: