Chapter 6: Problem 29
Locate the equilibrium point of the given nonhomogeneous linear system \(\mathbf{y}^{\prime}=A \mathbf{y}+\mathbf{g}_{0}\). [Hint: Introduce the change of dependent variable \(\mathbf{z}(t)=\mathbf{y}(t)-\mathbf{y}_{0}\), where \(\mathbf{y}_{0}\) is chosen so that the equation can be rewritten as \(\mathbf{z}^{\prime}=A \mathbf{z}\).] Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point. $$ \begin{aligned} &x^{\prime}=-x+2 \\ &y^{\prime}=2 y-4 \end{aligned} $$
Short Answer
Step by step solution
Set up the given nonhomogeneous linear system in matrix form
Apply the change of dependent variable
Determine \(\mathbf{y}_0\)
Rewrite the system with the new dependent variable
Classify the type and stability characteristics of the equilibrium point
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nonhomogeneous Linear System
This is represented in the format:
- \[ \mathbf{y}' = A\mathbf{y} + \mathbf{g}_0 \]
Matrix Form
For example, consider a system:
- \[ x' = -x + 2 \]
- \[ y' = 2y - 4 \]
- \[ \mathbf{y}' = A\mathbf{y} + \mathbf{g}_0 \]
Eigenvalues
- \[ \text{det}(A - \lambda I) = 0 \]
In the considered system, the eigenvalues for the matrix \( A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} \) are \( \lambda_1 = -1 \) and \( \lambda_2 = 2 \).
- Real eigenvalues: Indicate straightforward growth or decay along the corresponding direction.
- Complex eigenvalues would suggest oscillatory behavior.
Stability Classification
- Stable: If all eigenvalues are negative, any deviation from the equilibrium decays over time, returning to the equilibrium.
- Unstable: If at least one eigenvalue is positive, deviations from the equilibrium grow over time.
- Saddle Point: If eigenvalues have mixed signs, deviations in certain directions grow while others decay.