Math Problem Statement
Solution
The problem asks to solve a system of differential equations (DE) and analyze the behavior of the solution as . The given system of equations and initial conditions are:
with initial conditions:
Steps to Solve the System:
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Write the system in matrix form: where
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Find the eigenvalues and eigenvectors of :
- Compute the characteristic equation: where is the identity matrix. Expand to find .
- Solve for eigenvalues and corresponding eigenvectors.
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Express the solution: Using the eigenvalues and eigenvectors, the general solution is: where are eigenvectors, and are constants determined using initial conditions.
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Analyze the behavior as : The sign of the eigenvalues determines whether the solution grows, decays, or oscillates as .
Let me calculate the solution and the eigenvalues. numpy as np from scipy.linalg import eig
Define the matrix A
A = np.array([[4, 1], [2, 5]])
Compute eigenvalues and eigenvectors
eigenvalues, eigenvectors = eig(A)
Display the results
eigenvalues, eigenvectorsThe eigenvalues of the matrix are:
and the corresponding eigenvectors are:
\mathbf{v}_2 = \begin{bmatrix} -0.4472 \\ -0.8944 \end{bmatrix}.$$ ### General Solution The general solution to the system is: $$\mathbf{x}(t) = c_1 \mathbf{v}_1 e^{3t} + c_2 \mathbf{v}_2 e^{6t},$$ where $$c_1$$ and $$c_2$$ are determined from the initial conditions. Let me calculate $$c_1$$ and $$c_2$$ using the initial conditions. Define initial condition vector x0 = np.array([-2, 2]) # Solve for c1 and c2 using eigenvectors and initial conditions coefficients = np.linalg.solve(eigenvectors, x0) coefficientsThe coefficients $$c_1$$ and $$c_2$$ are: $$c_1 = 2.8284, \quad c_2 = 0.$$ ### Final Solution Substituting these values into the general solution, we get: $$\mathbf{x}(t) = 2.8284 \mathbf{v}_1 e^{3t},$$ where $$\mathbf{v}_1 = \begin{bmatrix} -0.7071 \\ 0.7071 \end{bmatrix}.$$ Expanding this, the solution is: $$x_1(t) = 2.8284 \cdot (-0.7071) e^{3t} = -2 e^{3t},$$ $$x_2(t) = 2.8284 \cdot 0.7071 e^{3t} = 2 e^{3t}.$$ ### Behavior as $$t \to \infty$$ Since the eigenvalue $$\lambda_1 = 3$$ is positive, the solution grows exponentially as $$t \to \infty$$. This indicates the system is unstable. Would you like additional details or have further questions? Here are some related queries: 1. How do eigenvalues influence system stability? 2. Can we solve similar systems with complex eigenvalues? 3. How does the behavior change if initial conditions differ? 4. Could we normalize the eigenvectors for simplicity? 5. How would adding a non-homogeneous term affect the solution? **Tip**: Always verify the stability of a system by analyzing the signs of eigenvalues!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Systems of Differential Equations
Matrix Algebra
Eigenvalues and Eigenvectors
Stability Analysis
Formulas
General Solution: \( \mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} \)
Characteristic Equation: \( \det(A - \lambda I) = 0 \)
Theorems
Stability Theorem using Eigenvalues
Superposition Principle for Linear Systems
Suitable Grade Level
Undergraduate (1st-2nd Year)
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