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The question provided is as follows:

**Q2:**
1. **(a)** Determine the type of critical point and its stability.
2. **(b)** Find the critical point for the system given by:

\[
\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -3 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.
\]

### Solution Outline

Let's address each part step-by-step.

### Part (a): Determine the Type and Stability of the Critical Point

1. **Identify the critical point**: For linear systems of the form \( \mathbf{x}' = A \mathbf{x} \), the origin \((0, 0)\) is typically a critical point.
  
2. **Find the Eigenvalues of the Matrix**: To classify the critical point and determine stability, we need to find the eigenvalues of the matrix
   \[
   A = \begin{pmatrix} 0 & 1 \\ -3 & -2 \end{pmatrix}.
   \]

   The eigenvalues \(\lambda\) are found by solving the characteristic polynomial:
   \[
   \det(A - \lambda I) = 0.
   \]
   This becomes:
   \[
   \det \begin{pmatrix} -\lambda & 1 \\ -3 & -2 - \lambda \end{pmatrix} = 0.
   \]

   Expanding the determinant:
   \[
   (-\lambda)(-2 - \lambda) - (1)(-3) = \lambda^2 + 2\lambda + 3 = 0.
   \]

   Solving this quadratic equation:
   \[
   \lambda = \frac{-2 \pm \sqrt{4 - 4 \cdot 3}}{2} = \frac{-2 \pm \sqrt{-8}}{2} = -1 \pm i\sqrt{2}.
   \]

3. **Classify the Eigenvalues**: Since the eigenvalues are complex, \(\lambda = -1 \pm i\sqrt{2}\), they indicate a **spiral** type of critical point. The negative real part \(-1\) means that the system is **asymptotically stable** (the origin is an **attracting spiral**).

### Part (b): Find the Critical Point

Since this is a linear homogeneous system, the only critical point is at the origin, \((0, 0)\).

### Summary

1. **Type of Critical Point**: Spiral (because of complex eigenvalues).
2. **Stability**: Asymptotically stable (because the real part of the eigenvalues is negative).

Would you like further explanation on any step, or details on eigenvalue calculation?

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Here are some related questions:

1. What does it mean for a system to be asymptotically stable?
2. How do complex eigenvalues affect the behavior of a dynamical system?
3. What types of critical points exist in dynamical systems?
4. How do we determine stability for non-linear systems?
5. What is the role of the Jacobian matrix in stability analysis?

**Tip:** Complex eigenvalues with a negative real part indicate an attracting spiral, which means trajectories will spiral towards the critical point.

Determine the type of critical point and its stability for the system given by: \( \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -3 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \).

This solution explains how to determine the type and stability of the critical point for a 2D linear system with a given matrix. By calculating eigenvalues, it classifies the critical point as a spiral type with asymptotic stability. Suitable for undergraduate math and engineering students.

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