Problem 28
Question
Use (1) to find the general solution of $$ \mathbf{X}^{\prime}=\left(\begin{array}{rrrr} -4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{array}\right) \mathbf{X} $$ Use a CAS to find \(e^{\mathrm{A} t}\).
Step-by-Step Solution
Verified Answer
Use a CAS to find \( e^{\mathbf{A} t} \) for the general solution.
1Step 1: Identify the Matrix A
The first step is to identify the matrix \( \mathbf{A} \) from the given differential equation \( \mathbf{X}' = \mathbf{A} \mathbf{X} \). The matrix \( \mathbf{A} \) is: \[\begin{pmatrix} -4 & 0 & 6 & 0 \0 & -5 & 0 & -4 \-1 & 0 & 1 & 0 \0 & 3 & 0 & 2 \end{pmatrix}\]
2Step 2: Use a Computer Algebra System (CAS)
To find \( e^{\mathbf{A} t} \), we use a Computer Algebra System (CAS) as manually computing the matrix exponential for a 4x4 matrix can be very complex and time-consuming. Input the matrix \( \mathbf{A} \) into the CAS and use the function to compute matrix exponentials, specifying the variable \( t \).
3Step 3: Analyze the Output
The CAS will output the matrix \( e^{\mathbf{A} t} \), which you can define as the general solution for the system \( \mathbf{X}(t) = e^{\mathbf{A} t} \mathbf{X}_0 \), where \( \mathbf{X}_0 \) is the initial state vector of the system.
Key Concepts
Matrix ExponentialComputer Algebra SystemGeneral Solution
Matrix Exponential
When dealing with matrix differential equations like \(\mathbf{X}^{\prime} = \mathbf{A}\mathbf{X}\), a crucial step in finding the solution is determining the matrix exponential, denoted as \(e^{\mathbf{A}t}\). This concept extends the idea of the exponential function to matrices. For a given square matrix \(\mathbf{A}\), \(e^{\mathbf{A}t}\) can be computed in a manner similar to the series expansion of the exponential function:
The matrix exponential is pivotal because it allows us to express the solution to the differential equation in the form \(\mathbf{X}(t) = e^{\mathbf{A}t}\mathbf{X}_0\), where \(\mathbf{X}_0\) is the initial condition.
- \(e^{\mathbf{A}t} = \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots\)
The matrix exponential is pivotal because it allows us to express the solution to the differential equation in the form \(\mathbf{X}(t) = e^{\mathbf{A}t}\mathbf{X}_0\), where \(\mathbf{X}_0\) is the initial condition.
Computer Algebra System
Utilizing a Computer Algebra System (CAS) becomes particularly valuable when you need to compute complex mathematical operations like matrix exponentials, which can be cumbersome and error-prone if done by hand. A CAS is a software tool that can quickly and accurately perform algebraic manipulations and solve algebraic expressions.
Therefore, it's recommended to familiarize yourself with a CAS for academic work that requires manipulating matrices and solving differential equations.
- It simplifies the calculation process by handling complex tasks like matrix exponentiation automatically.
- Users input the matrix and specify parameters like the variable \(t\).
- The CAS outputs results efficiently and without manual computation errors.
Therefore, it's recommended to familiarize yourself with a CAS for academic work that requires manipulating matrices and solving differential equations.
General Solution
After obtaining the matrix exponential \(e^{\mathbf{A}t}\), the next step is to find the general solution to the matrix differential equation \(\mathbf{X}^{\prime} = \mathbf{A}\mathbf{X}\). This general solution takes the form:
- \(\mathbf{X}(t) = e^{\mathbf{A}t} \mathbf{X}_0\)
- It describes how the system evolves over time from any given initial condition.
- It is applicable regardless of the specific initial values, making it a powerful tool in analyzing system dynamics under different scenarios.
Other exercises in this chapter
Problem 27
In Problems 21-30, find the general solution of the given system. $$ K^{\prime}=\left(\begin{array}{rrr} 5 & -4 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 5 \end{array}\right)
View solution Problem 28
Use variation of parameters to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right) \mathbf{x}+\left(\begin{ar
View solution Problem 28
Find the general solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & -1 & 1 \end{array}\right) \mathbf{X
View solution Problem 28
In Problems 13-32, use vaniation of parameters to solve the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right) \mat
View solution