Chapter 10: Problem 2
Show that \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) . \(\begin{array}{l}{A=\left[ \begin{array}{rr}{-2} & {1} \\ {1} & {-2}\end{array}\right]} \\ {x_{1}(t)=\frac{1}{2}\left(e^{-3 t}+e^{-1}\right),} & {x_{2}(t)=\frac{1}{2}\left(-e^{-3 t}+e^{-t}\right)}\end{array}\)
Short Answer
Step by step solution
Understand the System of Equations
Calculate the Derivatives of \(x_1(t)\) and \(x_2(t)\)
Express x in Vector Form
Perform Matrix Multiplication
Verify the Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Multiplication
For example, consider multiplying matrix \( A \) by a vector \( \mathbf{x}(t) \):
- Matrix \( A \) is given by \( \begin{bmatrix} -2 & 1 \ 1 & -2 \end{bmatrix} \).
- The vector \( \mathbf{x}(t) \) is \( \begin{bmatrix} \frac{1}{2}(e^{-3t} + e^{-t}) \ \frac{1}{2}(-e^{-3t} + e^{-t}) \end{bmatrix} \).
- First component: \(-2 \times \frac{1}{2}(e^{-3t} + e^{-t}) + 1 \times \frac{1}{2}(-e^{-3t} + e^{-t}) = \frac{1}{2}(-3e^{-3t} - e^{-t}) \).
- Second component: \(1 \times \frac{1}{2}(e^{-3t} + e^{-t}) - 2 \times \frac{1}{2}(-e^{-3t} + e^{-t}) = \frac{1}{2}(3e^{-3t} - e^{-t}) \).
System of Equations
In this scenario, the system of differential equations given is:\[\frac{d}{dt}\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} -2 & 1 \ 1 & -2 \end{bmatrix}\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\]
This indicates a relationship between derivatives \( \frac{d}{dt}x_1(t) \) and \( \frac{d}{dt}x_2(t) \) and the functions \( x_1(t) \) and \( x_2(t) \).
To solve this system, you find functions \( x_1(t) \) and \( x_2(t) \) that satisfy both equations simultaneously with their respective derivatives matching the transformation imposed by matrix \( A \). This is crucial in applications such as physics and engineering, where systems are modeled to predict behaviors and responses.
Vector Calculus
For vector-valued functions like \( \mathbf{x}(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \), the derivative \( \frac{d \mathbf{x}}{dt} \) represents the rate of change of both components \( x_1(t) \) and \( x_2(t) \) with respect to time \( t \).
- The derivative \( \frac{d}{dt} x_1(t) \) shows how the function \( x_1(t) \) changes over time.
- Similarly, \( \frac{d}{dt} x_2(t) \) provides insight into the temporal evolution of \( x_2(t) \).
- The notation \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) encapsulates the idea that the rate of change of the vector \( \mathbf{x} \) is influenced by both its current state and the dynamics described by \( A \).