Math Problem Statement

Solve the initial value problem (IVP) for the system of linear differential equations: y' = [[-11, 8], [-2, -3]]y, with initial condition y(0) = [6, 2].

Solution

The problem involves solving the initial value problem (IVP) for the system of differential equations given by:

y=[11823]yy' = \begin{bmatrix} -11 & 8 \\ -2 & -3 \end{bmatrix} y with initial condition: y(0)=[62]y(0) = \begin{bmatrix} 6 \\ 2 \end{bmatrix}

Solution Steps:

  1. Find the eigenvalues and eigenvectors of the matrix [11823]\begin{bmatrix} -11 & 8 \\ -2 & -3 \end{bmatrix}.
  2. Solve the system using diagonalization or other techniques based on the structure of the matrix.
  3. Use the initial condition y(0)=[62]y(0) = \begin{bmatrix} 6 \\ 2 \end{bmatrix} to find specific constants in the solution.

I'll begin by finding the eigenvalues and eigenvectors, which are crucial to solving this system. Let me calculate that.The eigenvalues of the matrix [11823]\begin{bmatrix} -11 & 8 \\ -2 & -3 \end{bmatrix} are both 7-7, indicating that the system has repeated eigenvalues.

The corresponding eigenvectors are:

[0.89440.4472]and[0.89440.4472]\begin{bmatrix} 0.8944 \\ 0.4472 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} -0.8944 \\ -0.4472 \end{bmatrix}

Given the repeated eigenvalue, the general solution to the system will take the form:

y(t)=c1e7t[0.89440.4472]+c2e7tt[0.89440.4472]y(t) = c_1 e^{-7t} \begin{bmatrix} 0.8944 \\ 0.4472 \end{bmatrix} + c_2 e^{-7t} t \begin{bmatrix} 0.8944 \\ 0.4472 \end{bmatrix}

Next, we'll determine the constants c1c_1 and c2c_2 using the initial condition y(0)=[62]y(0) = \begin{bmatrix} 6 \\ 2 \end{bmatrix}. Let me solve for these constants.It appears there was an error because I incorrectly attempted to solve the equation using a non-square matrix. I will correct the method and solve for the constants c1c_1 and c2c_2 using a different approach.

Let's set up the system of equations for c1c_1 and c2c_2 based on the initial condition, and solve it directly.The values of c1c_1 and c2c_2 are:

  • c1=6.26c_1 = 6.26
  • c2=[12.073.58]c_2 = \begin{bmatrix} -12.07 \\ 3.58 \end{bmatrix}

Thus, the solution to the system of differential equations is:

y(t)=6.26e7t[0.89440.4472]+t(12.07e7t[0.89440.4472]+3.58e7t[0.89440.4472])y(t) = 6.26 e^{-7t} \begin{bmatrix} 0.8944 \\ 0.4472 \end{bmatrix} + t \left( -12.07 e^{-7t} \begin{bmatrix} 0.8944 \\ 0.4472 \end{bmatrix} + 3.58 e^{-7t} \begin{bmatrix} 0.8944 \\ 0.4472 \end{bmatrix} \right)

If you have any questions or would like more details, feel free to ask!

Here are 5 related questions to explore:

  1. How can we handle repeated eigenvalues in systems of differential equations?
  2. What is the physical meaning of eigenvalues and eigenvectors in dynamic systems?
  3. How does diagonalization simplify solving systems of differential equations?
  4. Can we apply the same method to non-linear systems? How?
  5. What alternative methods exist for solving linear systems of differential equations?

Tip: Repeated eigenvalues often require special handling, such as generalized eigenvectors, when solving systems.

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Math Problem Analysis

Mathematical Concepts

Linear Systems of Differential Equations
Eigenvalues and Eigenvectors
Repeated Eigenvalues

Formulas

General solution for systems with repeated eigenvalues: y(t) = c1 e^(λt) v1 + c2 t e^(λt) v2
Eigenvalue equation: det(A - λI) = 0

Theorems

Eigenvalue and Eigenvector Theorem
Diagonalization of Matrices

Suitable Grade Level

Undergraduate level or Advanced High School (Calculus)