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