Math Problem Statement

For the system dY/dt = AY with A = [[2, 2], [-4, 6]], classify the equilibrium point at the origin.

Solution

To classify the equilibrium point of the system dYdt=AY\frac{dY}{dt} = AY with A=[2246]A = \begin{bmatrix} 2 & 2 \\ -4 & 6 \end{bmatrix}, we need to:

  1. Compute the eigenvalues of the matrix AA to determine the nature of the equilibrium point.

Step 1: Calculate the eigenvalues of AA

The eigenvalues λ\lambda of AA are found by solving the characteristic equation: det(AλI)=0\text{det}(A - \lambda I) = 0 where II is the identity matrix. The equation becomes: det[2λ246λ]=0\text{det} \begin{bmatrix} 2 - \lambda & 2 \\ -4 & 6 - \lambda \end{bmatrix} = 0

The determinant is: (2λ)(6λ)(4)(2)=0(2 - \lambda)(6 - \lambda) - (-4)(2) = 0 (2λ)(6λ)+8=0(2 - \lambda)(6 - \lambda) + 8 = 0 122λ6λ+λ2+8=012 - 2\lambda - 6\lambda + \lambda^2 + 8 = 0 λ28λ+20=0\lambda^2 - 8\lambda + 20 = 0


Step 2: Solve the quadratic equation

λ28λ+20=0\lambda^2 - 8\lambda + 20 = 0

Using the quadratic formula: λ=b±b24ac2a\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=8b = -8, and c=20c = 20, we find: λ=(8)±(8)24(1)(20)2(1)\lambda = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(20)}}{2(1)} λ=8±64802\lambda = \frac{8 \pm \sqrt{64 - 80}}{2} λ=8±162\lambda = \frac{8 \pm \sqrt{-16}}{2} λ=8±4i2\lambda = \frac{8 \pm 4i}{2} λ=4±2i\lambda = 4 \pm 2i


Step 3: Classify the equilibrium

The eigenvalues λ=4±2i\lambda = 4 \pm 2i are complex with a positive real part (Re(λ)=4\text{Re}(\lambda) = 4). This indicates that the equilibrium point at the origin is a spiral source.

The direction (clockwise or counterclockwise) is determined by the sign of the imaginary part (Im(λ)=2\text{Im}(\lambda) = 2):

  • Since Im(λ)>0\text{Im}(\lambda) > 0, the rotation is counterclockwise.

Final Answer:

The equilibrium point is a spiral source, counterclockwise.

Let me know if you'd like further clarification!


Related Questions:

  1. How do we interpret eigenvalues geometrically in 2D systems?
  2. Why does a positive real part indicate instability (source behavior)?
  3. What changes if the eigenvalues have a negative real part instead?
  4. Can a center equilibrium point occur with complex eigenvalues?
  5. How does the sign of the imaginary part determine rotation direction?

Tip: For 2D systems, always consider both real and imaginary parts of eigenvalues for classification.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Differential Equations
Eigenvalues and Eigenvectors

Formulas

Characteristic equation det(A - λI) = 0
Quadratic equation formula λ = (-b ± sqrt(b² - 4ac)) / 2a

Theorems

Stability analysis using eigenvalues

Suitable Grade Level

Undergraduate (Calculus and Linear Algebra level)