lowing values. ㅏ t) = 2t e 20et 35et x(0) = + 18- -t e [] eAt 3 4e2t-e-t 4e2t - 4e-t - e
lowing values. ㅏ t) = 2t e 20et 35et x(0) = + 18- -t e [] eAt 3 4e2t-e-t 4e2t - 4e-t - e
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
![**Variation of Parameters in Linear Differential Equations**
In this exercise, we explore solving the initial value problem using the method of variation of parameters. The problem is expressed by the differential equation \( \mathbf{x'} = A\mathbf{x} + \mathbf{f}(t) \), where \( \mathbf{x}(a) = \mathbf{x}_a \).
**Given Values:**
- \( A = \begin{bmatrix} 3 & -1 \\ 4 & -2 \end{bmatrix} \)
- \( \mathbf{f}(t) = \begin{bmatrix} 20e^t \\ 35e^t \end{bmatrix} \)
- Initial condition: \( \mathbf{x}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)
- The matrix exponential \( e^{At} \) is calculated as:
\[
e^{At} = \frac{1}{3} \begin{bmatrix} 4e^{2t} - e^{-t} & -e^{2t} + e^{-t} \\ 4e^{2t} - 4e^{-t} & -e^{2t} + 4e^{-t} \end{bmatrix}
\]
**Solution Representation:**
- The solution \( \mathbf{x}(t) \) of the differential equation is given by:
\[
\mathbf{x}(t) = \begin{bmatrix} e^{2t} + \boxed{\phantom{e}} \\ e^{2t} + \boxed{\phantom{e}} \end{bmatrix} e^{-t}
\]
**Explanation:**
- The initial segments of the matrices depict the setup using the matrix \( A \), function vector \( \mathbf{f}(t) \), and the exponentiation of matrix \( A \).
- The solution for \( \mathbf{x}(t) \) is expressed in matrix form, using exponential functions to account for the particular solution induced by \( \mathbf{f}(t) \).
In the diagram, blue squares represent placeholders for steps or intermediate results in the computation, emphasizing the structure of the final solution matrix. Adjusting these parameters is crucial for tailoring the solution to specific initial conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3200e892-7b83-4670-8aa5-c4c84f2a6adb%2F4c019453-b82c-4de8-b2db-9b02d203c14e%2Fbpxefio_processed.png&w=3840&q=75)
Transcribed Image Text:**Variation of Parameters in Linear Differential Equations**
In this exercise, we explore solving the initial value problem using the method of variation of parameters. The problem is expressed by the differential equation \( \mathbf{x'} = A\mathbf{x} + \mathbf{f}(t) \), where \( \mathbf{x}(a) = \mathbf{x}_a \).
**Given Values:**
- \( A = \begin{bmatrix} 3 & -1 \\ 4 & -2 \end{bmatrix} \)
- \( \mathbf{f}(t) = \begin{bmatrix} 20e^t \\ 35e^t \end{bmatrix} \)
- Initial condition: \( \mathbf{x}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)
- The matrix exponential \( e^{At} \) is calculated as:
\[
e^{At} = \frac{1}{3} \begin{bmatrix} 4e^{2t} - e^{-t} & -e^{2t} + e^{-t} \\ 4e^{2t} - 4e^{-t} & -e^{2t} + 4e^{-t} \end{bmatrix}
\]
**Solution Representation:**
- The solution \( \mathbf{x}(t) \) of the differential equation is given by:
\[
\mathbf{x}(t) = \begin{bmatrix} e^{2t} + \boxed{\phantom{e}} \\ e^{2t} + \boxed{\phantom{e}} \end{bmatrix} e^{-t}
\]
**Explanation:**
- The initial segments of the matrices depict the setup using the matrix \( A \), function vector \( \mathbf{f}(t) \), and the exponentiation of matrix \( A \).
- The solution for \( \mathbf{x}(t) \) is expressed in matrix form, using exponential functions to account for the particular solution induced by \( \mathbf{f}(t) \).
In the diagram, blue squares represent placeholders for steps or intermediate results in the computation, emphasizing the structure of the final solution matrix. Adjusting these parameters is crucial for tailoring the solution to specific initial conditions.
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