I1 = I2 || = Solve the following initial value problem: 20 ²-(-25) * + (2000) ¹) ² + (501), #(¹) = (22) 72 d dt =
I1 = I2 || = Solve the following initial value problem: 20 ²-(-25) * + (2000) ¹) ² + (501), #(¹) = (22) 72 d dt =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![### Initial Value Problem for Differential Equations
#### Problem Statement:
Solve the following initial value problem:
\[ \frac{d}{dt} \vec{x} = \begin{pmatrix} -5 & 1 \\ -25 & 5 \end{pmatrix} \vec{x} + \begin{pmatrix} 56t^6 \\ 20t^3 \end{pmatrix}, \quad \vec{x}(1) = \begin{pmatrix} 20 \\ 72 \end{pmatrix} \]
Where:
- \(\frac{d}{dt} \vec{x}\) represents the derivative of vector \(\vec{x}\) with respect to time \(t\).
- The matrix \(\begin{pmatrix} -5 & 1 \\ -25 & 5 \end{pmatrix}\) is the coefficient matrix.
- The term \(\begin{pmatrix} 56t^6 \\ 20t^3 \end{pmatrix}\) is a non-homogeneous part of the differential equation.
- The initial condition is provided as \(\vec{x}(1) = \begin{pmatrix} 20 \\ 72 \end{pmatrix}\).
#### Solution:
To find the functions \( x_1(t) \) and \( x_2(t) \) that satisfies the above system:
1. **Determine the Homogeneous Solution**:
- Find the eigenvalues and eigenvectors of the coefficient matrix \(\begin{pmatrix} -5 & 1 \\ -25 & 5 \end{pmatrix}\).
2. **Find the Particular Solution**:
- Determine a specific solution for the non-homogeneous term \(\begin{pmatrix} 56t^6 \\ 20t^3 \end{pmatrix}\).
3. **Combine the Solutions**:
- Combine both the homogeneous and particular solutions.
4. **Apply Initial Condition**:
- Use \(\vec{x}(1) = \begin{pmatrix} 20 \\ 72 \end{pmatrix}\) to resolve any constants.
#### Final Equation:
\[ x_1 = \]
\[ x_2 = \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F216015d9-daff-4e98-b3d1-7f3f5202e820%2Fa6c49e2b-aeb6-47f3-9d15-8ea9fbfae5d8%2Fvtuv7r_processed.png&w=3840&q=75)
Transcribed Image Text:### Initial Value Problem for Differential Equations
#### Problem Statement:
Solve the following initial value problem:
\[ \frac{d}{dt} \vec{x} = \begin{pmatrix} -5 & 1 \\ -25 & 5 \end{pmatrix} \vec{x} + \begin{pmatrix} 56t^6 \\ 20t^3 \end{pmatrix}, \quad \vec{x}(1) = \begin{pmatrix} 20 \\ 72 \end{pmatrix} \]
Where:
- \(\frac{d}{dt} \vec{x}\) represents the derivative of vector \(\vec{x}\) with respect to time \(t\).
- The matrix \(\begin{pmatrix} -5 & 1 \\ -25 & 5 \end{pmatrix}\) is the coefficient matrix.
- The term \(\begin{pmatrix} 56t^6 \\ 20t^3 \end{pmatrix}\) is a non-homogeneous part of the differential equation.
- The initial condition is provided as \(\vec{x}(1) = \begin{pmatrix} 20 \\ 72 \end{pmatrix}\).
#### Solution:
To find the functions \( x_1(t) \) and \( x_2(t) \) that satisfies the above system:
1. **Determine the Homogeneous Solution**:
- Find the eigenvalues and eigenvectors of the coefficient matrix \(\begin{pmatrix} -5 & 1 \\ -25 & 5 \end{pmatrix}\).
2. **Find the Particular Solution**:
- Determine a specific solution for the non-homogeneous term \(\begin{pmatrix} 56t^6 \\ 20t^3 \end{pmatrix}\).
3. **Combine the Solutions**:
- Combine both the homogeneous and particular solutions.
4. **Apply Initial Condition**:
- Use \(\vec{x}(1) = \begin{pmatrix} 20 \\ 72 \end{pmatrix}\) to resolve any constants.
#### Final Equation:
\[ x_1 = \]
\[ x_2 = \]
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