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4. Solve the system by using the matrix exponential 1 480 x' = 0 1 -2 -5 -4

4. Solve the system by using the matrix exponential 1 480 x' = 0 1 -2 -5 -4

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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### Solving Systems with Matrix Exponentials **Problem Statement:** 4. Solve the system by using the matrix exponential: $$ x' = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -5 & -4 \end{bmatrix} x $$ **Explanation:** This problem involves solving a system of linear differential equations using the concept of the matrix exponential. The system is given in the form: $$ x' = Ax $$ where \( A \) is the coefficient matrix: $$ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -5 & -4 \end{bmatrix}. $$ **Step-by-Step Solution Using Matrix Exponential:** 1. **Compute the Matrix Exponential:** The general solution for the system can be found using the matrix exponential \( e^{At} \). The solution is given by: $$ x(t) = e^{At} x(0) $$ where \( x(0) \) is the initial condition vector. 2. **Find Eigenvalues and Eigenvectors:** To compute the matrix exponential, we typically start by finding the eigenvalues and eigenvectors of \( A \). 3. **Diagonalize the Matrix (if possible):** If \( A \) is diagonalizable, we can write \( A = PDP^{-1} \), where \( D \) is a diagonal matrix containing the eigenvalues of \( A \), and \( P \) is the matrix of eigenvectors. 4. **Compute the Exponential of the Diagonal Matrix:** If \( A \) is diagonalizable: $$ e^{At} = Pe^{Dt}P^{-1} $$ where $$ e^{Dt} = \begin{bmatrix} e^{\lambda_1 t} & 0 & 0 \\ 0 & e^{\lambda_2 t} & 0 \\ 0 & 0 & e^{\lambda_3 t} \end{bmatrix} $$ Here, \( \lambda_1, \lambda_2, \lambda_3 \) are the eigenvalues of \( A \). 5.
Transcribed Image Text:### Solving Systems with Matrix Exponentials **Problem Statement:** 4. Solve the system by using the matrix exponential: $$ x' = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -5 & -4 \end{bmatrix} x $$ **Explanation:** This problem involves solving a system of linear differential equations using the concept of the matrix exponential. The system is given in the form: $$ x' = Ax $$ where \( A \) is the coefficient matrix: $$ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -5 & -4 \end{bmatrix}. $$ **Step-by-Step Solution Using Matrix Exponential:** 1. **Compute the Matrix Exponential:** The general solution for the system can be found using the matrix exponential \( e^{At} \). The solution is given by: $$ x(t) = e^{At} x(0) $$ where \( x(0) \) is the initial condition vector. 2. **Find Eigenvalues and Eigenvectors:** To compute the matrix exponential, we typically start by finding the eigenvalues and eigenvectors of \( A \). 3. **Diagonalize the Matrix (if possible):** If \( A \) is diagonalizable, we can write \( A = PDP^{-1} \), where \( D \) is a diagonal matrix containing the eigenvalues of \( A \), and \( P \) is the matrix of eigenvectors. 4. **Compute the Exponential of the Diagonal Matrix:** If \( A \) is diagonalizable: $$ e^{At} = Pe^{Dt}P^{-1} $$ where $$ e^{Dt} = \begin{bmatrix} e^{\lambda_1 t} & 0 & 0 \\ 0 & e^{\lambda_2 t} & 0 \\ 0 & 0 & e^{\lambda_3 t} \end{bmatrix} $$ Here, \( \lambda_1, \lambda_2, \lambda_3 \) are the eigenvalues of \( A \). 5.
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