Math Problem Statement

Classify the equilibrium point (-1, -3) for the system of differential equations dx/dt = 10 - x^2 - y^2 and dy/dt = 3x - y.

Solution

To classify the equilibrium point (−1,−3)(-1, -3) for the given system of differential equations:

dxdt=10−x2−y2,\frac{dx}{dt} = 10 - x^2 - y^2, dydt=3x−y,\frac{dy}{dt} = 3x - y,

we need to follow these steps:

Step 1: Verify the Equilibrium Point

An equilibrium point satisfies dxdt=0\frac{dx}{dt} = 0 and dydt=0\frac{dy}{dt} = 0. Substituting x=−1x = -1 and y=−3y = -3:

dxdt=10−(−1)2−(−3)2=10−1−9=0,\frac{dx}{dt} = 10 - (-1)^2 - (-3)^2 = 10 - 1 - 9 = 0, dydt=3(−1)−(−3)=−3+3=0.\frac{dy}{dt} = 3(-1) - (-3) = -3 + 3 = 0.

Thus, (−1,−3)(-1, -3) is indeed an equilibrium point.

Step 2: Linearize the System

Compute the Jacobian matrix of the system:

\begin{bmatrix} \frac{\partial}{\partial x}\left(10 - x^2 - y^2\right) & \frac{\partial}{\partial y}\left(10 - x^2 - y^2\right) \\ \frac{\partial}{\partial x}\left(3x - y\right) & \frac{\partial}{\partial y}\left(3x - y\right) \end{bmatrix}.$$ Calculate each partial derivative: - $$\frac{\partial}{\partial x}\left(10 - x^2 - y^2\right) = -2x$$, - $$\frac{\partial}{\partial y}\left(10 - x^2 - y^2\right) = -2y$$, - $$\frac{\partial}{\partial x}\left(3x - y\right) = 3$$, - $$\frac{\partial}{\partial y}\left(3x - y\right) = -1$$. Thus, the Jacobian matrix is: $$J(x, y) = \begin{bmatrix} -2x & -2y \\ 3 & -1 \end{bmatrix}.$$ At the equilibrium point $$(-1, -3)$$: $$J(-1, -3) = \begin{bmatrix} -2(-1) & -2(-3) \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 3 & -1 \end{bmatrix}.$$ --- ### Step 3: Analyze the Eigenvalues of the Jacobian Find the eigenvalues $$\lambda$$ of the matrix $$J(-1, -3)$$ by solving $$\det(J - \lambda I) = 0$$: $$\det \begin{bmatrix} 2 - \lambda & 6 \\ 3 & -1 - \lambda \end{bmatrix} = 0.$$ Expanding the determinant: $$(2 - \lambda)(-1 - \lambda) - (6)(3) = 0,$$ $$(-2 - \lambda + \lambda^2) - 18 = 0,$$ $$\lambda^2 - \lambda - 20 = 0.$$ Solve this quadratic equation using the quadratic formula: $$\lambda = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)},$$ $$\lambda = \frac{1 \pm \sqrt{1 + 80}}{2},$$ $$\lambda = \frac{1 \pm \sqrt{81}}{2},$$ $$\lambda = \frac{1 \pm 9}{2}.$$ Thus: $$\lambda_1 = \frac{1 + 9}{2} = 5, \quad \lambda_2 = \frac{1 - 9}{2} = -4.$$ --- ### Step 4: Classify the Equilibrium Point Since the eigenvalues $$\lambda_1 = 5$$ and $$\lambda_2 = -4$$ have opposite signs, the equilibrium point is a **saddle point**. --- ### Final Answer: The correct classification is **Saddle point**. --- Would you like a detailed explanation of any step? ### Related Questions: 1. What is the significance of the Jacobian matrix in stability analysis? 2. How do the signs of eigenvalues determine the behavior of equilibrium points? 3. What does it mean for an equilibrium point to be a "spiral sink"? 4. How would the analysis change if both eigenvalues were positive? 5. Can you determine stability without explicitly solving for eigenvalues? ### Tip: When solving stability problems, always verify the equilibrium point first before proceeding with linearization.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differential Equations
Equilibrium Points
Jacobian Matrix
Eigenvalues

Formulas

dx/dt = 10 - x^2 - y^2
dy/dt = 3x - y
Jacobian Matrix: J(x, y) = [∂(dx/dt)/∂x ∂(dx/dt)/∂y; ∂(dy/dt)/∂x ∂(dy/dt)/∂y]
Eigenvalues: det(J - λI) = 0

Theorems

Linearization of dynamical systems
Stability analysis using eigenvalues

Suitable Grade Level

Grades 11-12

Related Recommendation

Classifying a Critical Point in a System of Differential Equations

Jacobian Matrix for the System dx/dt = 10 - x^2 - y^2, dy/dt = 3x - y at (1, 3)

Solving the Characteristic Equation for a Jacobian Matrix at Equilibrium

Analyzing Eigenvalues and Equilibrium Points in Direction Fields

Find Critical Points and Solve System of Differential Equations