5. Solve the following linear system: d..X dit with the initial condition - [83] X (0) = [3] 2 X
5. Solve the following linear system: d..X dit with the initial condition - [83] X (0) = [3] 2 X
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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![Title: Solving Linear Systems of Differential Equations
### Problem Statement
Solve the following linear system:
\[ \frac{d\mathbf{X}}{dt} = \begin{bmatrix} 3 & -2 \\ 0 & 3 \end{bmatrix} \mathbf{X} \]
with the initial condition
\[ \mathbf{X}(0) = \begin{bmatrix} -3 \\ 2 \end{bmatrix} \]
### Explanation
This problem involves solving a system of differential equations. The matrix equation provided can be written in the form of:
\[ \frac{d\mathbf{X}}{dt} = A\mathbf{X} \]
where
\[ A = \begin{bmatrix} 3 & -2 \\ 0 & 3 \end{bmatrix}, \]
\[ \mathbf{X} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. \]
### Steps to Solve:
1. **Find the Eigenvalues of \(A\):**
The eigenvalues, \(\lambda\), are found from the characteristic equation:
\[
\text{det}(A - \lambda I) = 0
\]
For the given matrix:
\[
A - \lambda I = \begin{bmatrix} 3 - \lambda & -2 \\ 0 & 3 - \lambda \end{bmatrix}
\]
The characteristic polynomial is:
\[
(3 - \lambda)^2 = 0
\]
Solving this gives:
\[
\lambda = 3 \quad \text{(with multiplicity 2)}
\]
2. **Find the Eigenvectors:**
For \(\lambda = 3\), solve:
\[
(A - 3I)\mathbf{v} = \mathbf{0}
\]
This simplifies to:
\[
\begin{bmatrix} 0 & -2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
\]
From this, \(v_2\) can be any arbitrary value, and \(v_1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff12b43ca-6b55-4575-a2c5-3f890463f057%2Fb5491541-419d-4397-bac0-441d2d5d18f0%2Ftn8xuvu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Title: Solving Linear Systems of Differential Equations
### Problem Statement
Solve the following linear system:
\[ \frac{d\mathbf{X}}{dt} = \begin{bmatrix} 3 & -2 \\ 0 & 3 \end{bmatrix} \mathbf{X} \]
with the initial condition
\[ \mathbf{X}(0) = \begin{bmatrix} -3 \\ 2 \end{bmatrix} \]
### Explanation
This problem involves solving a system of differential equations. The matrix equation provided can be written in the form of:
\[ \frac{d\mathbf{X}}{dt} = A\mathbf{X} \]
where
\[ A = \begin{bmatrix} 3 & -2 \\ 0 & 3 \end{bmatrix}, \]
\[ \mathbf{X} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. \]
### Steps to Solve:
1. **Find the Eigenvalues of \(A\):**
The eigenvalues, \(\lambda\), are found from the characteristic equation:
\[
\text{det}(A - \lambda I) = 0
\]
For the given matrix:
\[
A - \lambda I = \begin{bmatrix} 3 - \lambda & -2 \\ 0 & 3 - \lambda \end{bmatrix}
\]
The characteristic polynomial is:
\[
(3 - \lambda)^2 = 0
\]
Solving this gives:
\[
\lambda = 3 \quad \text{(with multiplicity 2)}
\]
2. **Find the Eigenvectors:**
For \(\lambda = 3\), solve:
\[
(A - 3I)\mathbf{v} = \mathbf{0}
\]
This simplifies to:
\[
\begin{bmatrix} 0 & -2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
\]
From this, \(v_2\) can be any arbitrary value, and \(v_1
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