x1 = I2 = Solve the following initial value problem: d dt (² 25 -5 -5) x² (101² 1 x + 2t -1 6), 10t ¹ +26 (1) = 13 + 65 T 1 26
x1 = I2 = Solve the following initial value problem: d dt (² 25 -5 -5) x² (101² 1 x + 2t -1 6), 10t ¹ +26 (1) = 13 + 65 T 1 26
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 an Initial Value Problem
#### Problem Statement:
Solve the following initial value problem:
\[ \frac{d}{dt} \vec{x} = \begin{pmatrix} 25 & -5 \\ -5 & 1 \end{pmatrix} \vec{x} + \begin{pmatrix} \frac{2t^{-1}}{10t^{-1} + 26} \\ \end{pmatrix}, \quad \vec{x}(1) = \begin{pmatrix} 13 + \frac{5}{26} \\ 65 - \frac{1}{26} \end{pmatrix} \]
---
#### Explanation:
- \(\frac{d}{dt} \vec{x}\): This denotes the derivative of the vector \(\vec{x}(t)\) with respect to \(t\).
- \(\begin{pmatrix} 25 & -5 \\ -5 & 1 \end{pmatrix}\): This is the coefficient matrix for the system of differential equations.
- \(\vec{x}\): This is the vector function of \(t\) that we need to find.
- \(\begin{pmatrix} \frac{2t^{-1}}{10t^{-1} + 26} \\ \end{pmatrix}\): This is the non-homogeneous term vector.
- \(\vec{x}(1) = \begin{pmatrix} 13 + \frac{5}{26} \\ 65 - \frac{1}{26} \end{pmatrix}\): This represents the initial condition given at \(t = 1\).
#### Solution:
First, let's rewrite and simplify the initial condition:
\[
\vec{x}(1) = \begin{pmatrix} 13 + \frac{5}{26} \\ 65 - \frac{1}{26} \end{pmatrix}
\]
Next, we solve the system of differential equations by firstly calculating the eigenvalues and eigenvectors of the coefficient matrix \(\begin{pmatrix} 25 & -5 \\ -5 & 1 \end{pmatrix}\), and then use them to solve the associated homogenous system.
After solving, we need to find the particular solution to incorporate the non-homogeneous part \(\begin{pmatrix} \frac{2t^{-1}}{10t^{-1} + 26} \\ \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F216015d9-daff-4e98-b3d1-7f3f5202e820%2F0bdb4c5e-486d-4856-8bae-0f564ed1f689%2Fco13wtl_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Solving an Initial Value Problem
#### Problem Statement:
Solve the following initial value problem:
\[ \frac{d}{dt} \vec{x} = \begin{pmatrix} 25 & -5 \\ -5 & 1 \end{pmatrix} \vec{x} + \begin{pmatrix} \frac{2t^{-1}}{10t^{-1} + 26} \\ \end{pmatrix}, \quad \vec{x}(1) = \begin{pmatrix} 13 + \frac{5}{26} \\ 65 - \frac{1}{26} \end{pmatrix} \]
---
#### Explanation:
- \(\frac{d}{dt} \vec{x}\): This denotes the derivative of the vector \(\vec{x}(t)\) with respect to \(t\).
- \(\begin{pmatrix} 25 & -5 \\ -5 & 1 \end{pmatrix}\): This is the coefficient matrix for the system of differential equations.
- \(\vec{x}\): This is the vector function of \(t\) that we need to find.
- \(\begin{pmatrix} \frac{2t^{-1}}{10t^{-1} + 26} \\ \end{pmatrix}\): This is the non-homogeneous term vector.
- \(\vec{x}(1) = \begin{pmatrix} 13 + \frac{5}{26} \\ 65 - \frac{1}{26} \end{pmatrix}\): This represents the initial condition given at \(t = 1\).
#### Solution:
First, let's rewrite and simplify the initial condition:
\[
\vec{x}(1) = \begin{pmatrix} 13 + \frac{5}{26} \\ 65 - \frac{1}{26} \end{pmatrix}
\]
Next, we solve the system of differential equations by firstly calculating the eigenvalues and eigenvectors of the coefficient matrix \(\begin{pmatrix} 25 & -5 \\ -5 & 1 \end{pmatrix}\), and then use them to solve the associated homogenous system.
After solving, we need to find the particular solution to incorporate the non-homogeneous part \(\begin{pmatrix} \frac{2t^{-1}}{10t^{-1} + 26} \\ \
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