Question

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 6 & -10 \\ 2 & -3 \end{array}\right] \mathbf{y} $$

Short answer

Question: Determine the type and stability of the equilibrium point of the following linear system of two differential equations and whether it is a proper or improper node: $$ \dot{x} = 6x - 10y \\ \dot{y} = 2x - 3y $$ Answer: The equilibrium point at the phase-plane origin for the given linear system of two differential equations is an unstable spiral. It is not a node, so the classification of the equilibrium point as a proper or improper node does not apply in this case.

Step-by-step solution

Compute the eigenvalues of matrix A

We need to find the eigenvalues of the given matrix A: $$ A = \begin{bmatrix} 6 & -10 \\ 2 & -3 \end{bmatrix} $$ To do that, we first find the determinant of \((A-\lambda I)\) where \(\lambda\) represents eigenvalues and \(I\) is the identity matrix. We will set the determinant equal to zero and solve for \(\lambda\). $$ \text{det}(A-\lambda I) = \text{det} \begin{bmatrix} 6-\lambda & -10 \\ 2 & -3 - \lambda \end{bmatrix}= (6-\lambda)(-3-\lambda) - (-10)(2) = \lambda^2 - 3\lambda - 4\lambda +18 +20 = 0 $$ Now we have a quadratic equation to solve for the eigenvalues: $$ \lambda^2 - 7\lambda + 38 = 0 $$

Solve the quadratic equation for λ

To find the eigenvalues, we can use the quadratic equation formula: $$ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ In this case, \(a=1\), \(b=-7\), and \(c=38\). Plug these values into the formula and solve for \(\lambda\). $$ \lambda = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(38)}}{2(1)} = \frac{7 \pm \sqrt{49 - 152}}{2} $$ Since the discriminant is negative, the eigenvalues are a complex conjugate pair. We can write them as: $$ \lambda_{1,2} = 3.5 \pm 0.5i $$

Classify the type and stability of the equilibrium point

Referring to Table 6.2 (classification table for equilibrium points), since we have complex conjugate eigenvalues with positive real parts, the equilibrium point at the phase-plane origin is an *unstable spiral*.

Determine if the equilibrium point is a proper or improper node

Since the equilibrium point is an unstable spiral, it is not a node. Thus, the classification of the equilibrium point as a proper or improper node does not apply in this case.

Key concepts

Eigenvalues

Eigenvalues are a fundamental concept when it comes to understanding linear systems. When analyzing a matrix, they give us valuable insights into the system's behavior. To find them for a given matrix, such as our matrix \[ A = \begin{bmatrix} 6 & -10 \ 2 & -3 \end{bmatrix} \], we first need to subtract \( \lambda \) from the diagonal elements, where \( \lambda \) represents the eigenvalues. - The next step is to compute the determinant of \( A - \lambda I \), which for our matrix becomes: \[\text{det}(A-\lambda I) = \text{det} \begin{bmatrix} 6-\lambda & -10 \ 2 & -3-\lambda \end{bmatrix} = (6-\lambda)(-3-\lambda) - (-10)(2).\] - Solving the resulting equation, \( \lambda^2 - 7\lambda + 38 = 0 \), reveals the eigenvalues. Here, the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) yields the solutions. Because the discriminant is negative, the solutions are complex, ending up with \( \lambda_{1,2} = 3.5 \pm 0.5i \).These complex eigenvalues suggest a specific type of dynamical behavior in the associated linear system.

Equilibrium Points

For a linear system, an equilibrium point is where the system does not change, or in other words, where the derivative \( \mathbf{y}' \) becomes zero. This means the system reaches a state of balance. In our specific problem, \( \mathbf{y} = \mathbf{0} \) is the sole equilibrium point due to the system's linear nature.- Analyzing the matrix helps us understand this point better. The matrix \( A \) describes how the system evolves over time.- Equilibrium points are crucial since they dictate how the solutions to the differential equation \( \mathbf{y}' = A\mathbf{y} \) behave near these points.By understanding these points, we begin to foresee the eventual paths or tendencies of any given trajectory within the system.

Phase-plane Analysis

Phase-plane analysis is a visual representation technique used to study differential equations, particularly in dynamic systems. In a 2D system, the phase plane shows us how the system evolves over time, represented as vectors or trajectories on a plane.- In our case, by examining the eigenvalues \( \lambda_{1,2} = 3.5 \pm 0.5i \), we can predict that the trajectories will form spirals around the equilibrium point.- Since the real part of the eigenvalues is positive (3.5), these spirals spiral outwards, indicating they move away from the equilibrium point, hinting at an unstable system.Thus, by plotting these trajectories, the phase-plane allows students and researchers to visually appreciate the system's behavior and derive insights from it.

Stability Analysis

Stability analysis helps determine whether an equilibrium point is stable or unstable. It informs how systems respond to small perturbations. In systems where disturbances can alter the course of the state, understanding this is crucial.- The eigenvalues \( \lambda_{1,2} = 3.5 \pm 0.5i \) are crucial because their real part signifies an important stability characteristic. - If the real part is positive, as in our case (3.5), it suggests that even small disturbances will cause the system to diverge from the equilibrium point.Therefore, the equilibrium point is classified as an *unstable spiral*. This implies that the trajectories tend to spiral away rather than settle into a stable configuration. Understanding this concept is vital for fields including engineering and physics, where predicting system behavior under small disruptions is often essential.