Math Problem Statement

Determine the type of critical point and stability of the system: \( \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -3 & -c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \)

Solution

The problem provided is as follows:

Q2:
(a) Determine the type of critical point and stability of
(xy)=(013c)(xy).\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -3 & -c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.

This is a problem related to finding the stability and type of critical points for a linear system of differential equations.

Solution Outline

To determine the type and stability of the critical point, follow these steps:

  1. Set Up the Characteristic Equation:
    The characteristic equation of a matrix A=(013c)A = \begin{pmatrix} 0 & 1 \\ -3 & -c \end{pmatrix} is given by det(AλI)=0\det(A - \lambda I) = 0.

  2. Compute the Determinant:
    For AλIA - \lambda I, we get: AλI=(λ13cλ).A - \lambda I = \begin{pmatrix} -\lambda & 1 \\ -3 & -c - \lambda \end{pmatrix}. The determinant is: det(AλI)=(λ)(cλ)(1)(3)=λ2+cλ+3=0.\det(A - \lambda I) = (-\lambda)(-c - \lambda) - (1)(-3) = \lambda^2 + c\lambda + 3 = 0.

  3. Solve for Eigenvalues:
    The eigenvalues, λ\lambda, are the solutions to the characteristic equation: λ2+cλ+3=0.\lambda^2 + c\lambda + 3 = 0. This quadratic equation in λ\lambda has roots: λ=c±c2432=c±c2122.\lambda = \frac{-c \pm \sqrt{c^2 - 4 \cdot 3}}{2} = \frac{-c \pm \sqrt{c^2 - 12}}{2}. The nature of these roots (real, complex, etc.) depends on the discriminant c212c^2 - 12.

  4. Determine the Type and Stability Based on Eigenvalues:

    • If c212>0c^2 - 12 > 0, the roots are real and distinct, leading to different stability behaviors depending on their sign.
    • If c212=0c^2 - 12 = 0, the roots are real and repeated.
    • If c212<0c^2 - 12 < 0, the roots are complex, indicating a spiral or center-type critical point.

The stability of the critical point depends on the real parts of the eigenvalues:

  • Stable Node or Spiral Sink if both real parts are negative.
  • Unstable Node or Spiral Source if both real parts are positive.
  • Saddle Point if the eigenvalues have opposite signs.

Would you like me to proceed with a specific value of cc or further analyze the general case? Additionally, if you have questions on each of these steps or need additional information on critical point classifications, let me know.


Here are some follow-up questions to deepen understanding:

  1. What happens to the critical point's stability when c=0c = 0?
  2. How does the type of critical point change as cc varies across positive and negative values?
  3. Why does the sign of the real part of the eigenvalues affect stability?
  4. How would you classify the critical point if the eigenvalues are complex with zero real parts?
  5. What physical systems could this type of differential equation matrix represent?

Tip: When analyzing stability, always pay close attention to the sign of the real part of the eigenvalues—this will determine if trajectories converge to or diverge from the critical point.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Eigenvalues and Eigenvectors
Stability Analysis

Formulas

Characteristic equation of a matrix A: \( \det(A - \lambda I) = 0 \)
Eigenvalues: \( \lambda = \frac{-c \pm \sqrt{c^2 - 12}}{2} \)

Theorems

Stability of linear systems
Classification of critical points

Suitable Grade Level

College