The critical point $( 0,0 )$ of the system $\frac { d x } { d t } = 4 x - y , \frac { d y } { d t } = x + 2 y$ is Select all that apply.
A) asymptotically stable
B) stable but not asymptotically stable
C) unstable
D) an attractor
E) a repeller

Question

Quizplus · Accepted Answer

The stability of a critical point can be determined by analyzing the eigenvalues of the Jacobian matrix of the system at that point. The Jacobian matrix for this system is:$$J = \begin{pmatrix}4 & -1 \1 & 2\end{pmatrix}$$The eigenvalues $\lambda$ are found by solving the characteristic equation $\det(J - \lambda I) = 0$, where $I$ is the identity matrix. This gives:$$\det\begin{pmatrix}4 - \lambda & -1 \1 & 2 - \lambda\end{pmatrix} = 0$$$$(4 - \lambda)(2 - \lambda) - (-1)(1) = \lambda^2 - 6\lambda + 9 - 1 = \lambda^2 - 6\lambda + 8 = 0$$$$\lambda^2 - 6\lambda + 8 = (\lambda - 4)(\lambda - 2) = 0$$So, the eigenvalues are $\lambda_1 = 4$ and $\lambda_2 = 2$, both of which are positive.A critical point is unstable if any of the eigenvalues of the Jacobian matrix at that point have positive real parts, which is the case here. Therefore, the critical point $(0,0)$ is unstable.An attractor is a type of stable critical point that draws nearby trajectories towards itself, while a repeller is an unstable critical point that pushes nearby trajectories away. Since the critical point in this case is unstable and has positive eigenvalues, it acts as a repeller, pushing trajectories away from itself.

The Critical Point (0,0)( 0,0 )(0,0) Of the System dxdt=4x−y,dydt=x+2y\frac { d x } { d t } = 4 x - y , \frac { d y } { d t } = x + 2 ydtdx​=4x−y,dtdy​=x+2y

The Critical Point $( 0,0 )$ Of the System $\frac { d x } { d t } = 4 x - y , \frac { d y } { d t } = x + 2 y$