Chapter 8: Problem 7
Use the method of undetermined coefficients to solve the given non-homogeneous system. $$\mathbf{X}^{\prime}=\left(\begin{array}{lll} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 5 \end{array}\right) \mathbf{X}+\left(\begin{array}{r} 1 \\ -1 \\ 2 \end{array}\right) e^{4 t}$$
Short Answer
Expert verified
General solution: Combine homogeneous and particular solutions. Solve system for coefficients.
Step by step solution
01
Find the Homogeneous Solution
Start by solving the homogeneous system \( \mathbf{X}^{\prime} = A \mathbf{X} \), where \( A = \begin{pmatrix} 1 & 1 & 1 \ 0 & 2 & 3 \ 0 & 0 & 5 \end{pmatrix} \). The characteristic equation is determined by the eigenvalues of matrix \( A \), which are the diagonal elements: \( \lambda_1 = 1 \), \( \lambda_2 = 2 \), and \( \lambda_3 = 5 \). The general solution for the homogeneous system is \( \mathbf{X}_h = c_1 \mathbf{v}_1 e^{t} + c_2 \mathbf{v}_2 e^{2t} + c_3 \mathbf{v}_3 e^{5t} \), where \( c_1, c_2, \) and \( c_3 \) are constants and \( \mathbf{v}_1, \mathbf{v}_2, \) and \( \mathbf{v}_3 \) are the eigenvectors corresponding to the eigenvalues.
02
Formulate the Particular Solution
Seek a particular solution \( \mathbf{X}_p \) of the form \( \mathbf{X}_p = \mathbf{a} e^{4t} \), where \( \mathbf{a} = \begin{pmatrix} a_1 \ a_2 \ a_3 \end{pmatrix} \) is a vector to be determined. Plug this form into the non-homogeneous equation: \( \mathbf{X}^{\prime} = A \mathbf{X} + \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} e^{4t} \). This yields \( 4\mathbf{a} e^{4t} = A\mathbf{a} e^{4t} + \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} e^{4t} \).
03
Solve for Coefficients of Particular Solution
Equating coefficients of \( e^{4t} \), solve the equation \( (4I - A) \mathbf{a} = \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} \), where \( I \) is the identity matrix. Plug in the matrices to obtain: \( \begin{pmatrix} 3 & -1 & -1 \ 0 & 2 & -3 \ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} a_1 \ a_2 \ a_3 \end{pmatrix} = \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} \). Solve this system of linear equations to find \( a_1, a_2, \) and \( a_3 \).
04
Combine Solutions for General Solution
Once \( \mathbf{a} = \begin{pmatrix} a_1 \ a_2 \ a_3 \end{pmatrix} \) is found, the general solution is \( \mathbf{X} = \mathbf{X}_h + \mathbf{X}_p \). Substitute \( \mathbf{X}_h = c_1 \mathbf{v}_1 e^{t} + c_2 \mathbf{v}_2 e^{2t} + c_3 \mathbf{v}_3 e^{5t} \) and \( \mathbf{X}_p = \mathbf{a} e^{4t} \) into \( \mathbf{X} \). The result is the general solution to the non-homogeneous system.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Non-Homogeneous System
A non-homogeneous system of differential equations is one where the system includes a non-zero term on the right-hand side of the equation. In mathematical terms, it looks like this: \( \mathbf{X}' = A\mathbf{X} + \mathbf{F}(t) \). Here, \( A\mathbf{X} \) represents the homogeneous part, while \( \mathbf{F}(t) \) is the non-homogeneous term, often a function of time, \( t \).
Non-homogeneous systems are more complex than homogeneous ones because they include an extra function. This makes them slightly harder to solve, as they require finding both a homogeneous solution and a particular solution. When combined, these solutions give the general solution to the system.
Understanding non-homogeneous systems is crucial because they often model real-world phenomena, where external forces or influences (represented by the non-homogeneous term) act on the system.
Non-homogeneous systems are more complex than homogeneous ones because they include an extra function. This makes them slightly harder to solve, as they require finding both a homogeneous solution and a particular solution. When combined, these solutions give the general solution to the system.
Understanding non-homogeneous systems is crucial because they often model real-world phenomena, where external forces or influences (represented by the non-homogeneous term) act on the system.
Homogeneous Solution
The homogeneous solution is found by solving the system \( \mathbf{X}' = A\mathbf{X} \), where you ignore any non-homogeneous terms. This involves determining the characteristic equation of the matrix \( A \) to find the eigenvalues.
Once you have the eigenvalues, \( \lambda_1, \lambda_2, \) and \( \lambda_3 \), you can find the corresponding eigenvectors, \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \).
The homogeneous solution takes the form \( \mathbf{X}_h = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + c_3 \mathbf{v}_3 e^{\lambda_3 t} \).
Once you have the eigenvalues, \( \lambda_1, \lambda_2, \) and \( \lambda_3 \), you can find the corresponding eigenvectors, \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \).
The homogeneous solution takes the form \( \mathbf{X}_h = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + c_3 \mathbf{v}_3 e^{\lambda_3 t} \).
- \( c_1, c_2, \) and \( c_3 \) are arbitrary constants.
- Each term is a product of an exponential function and an eigenvector, reflecting how the states change over time independently according to each eigenvalue.
Particular Solution
The particular solution, \( \mathbf{X}_p \), accounts for the non-homogeneous part of the system, \( \mathbf{F}(t) \). In the method of undetermined coefficients, you assume a form for \( \mathbf{X}_p \) that you think matches \( \mathbf{F}(t) \).
In our example, \( \mathbf{F}(t) = \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} e^{4t} \). Thus, \( \mathbf{X}_p \) is assumed to take a similar exponential form \( \mathbf{X}_p = \mathbf{a} e^{4t} \), where \( \mathbf{a} \) is a vector.
To determine \( \mathbf{a} \), substitute \( \mathbf{X}_p \) back into the differential equation. This involves calculating \( \mathbf{X}' = A\mathbf{X}_p + \mathbf{F}(t) \), then solving for \( \mathbf{a} \) by equating coefficients of \( e^{4t} \).
In our example, \( \mathbf{F}(t) = \begin{pmatrix} 1 \ -1 \ 2 \end{pmatrix} e^{4t} \). Thus, \( \mathbf{X}_p \) is assumed to take a similar exponential form \( \mathbf{X}_p = \mathbf{a} e^{4t} \), where \( \mathbf{a} \) is a vector.
To determine \( \mathbf{a} \), substitute \( \mathbf{X}_p \) back into the differential equation. This involves calculating \( \mathbf{X}' = A\mathbf{X}_p + \mathbf{F}(t) \), then solving for \( \mathbf{a} \) by equating coefficients of \( e^{4t} \).
- This step typically results in a system of equations, which you solve to find the components of \( \mathbf{a} \).
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental concepts in linear algebra and play a crucial role in solving systems of differential equations. An eigenvalue \( \lambda \) is a scalar that indicates how much and in what direction a transformation stretches a vector.
For a matrix \( A \), an eigenvector \( \mathbf{v} \) satisfies the equation \( A\mathbf{v} = \lambda\mathbf{v} \). Here, \( \lambda \) is the eigenvalue corresponding to this eigenvector.
In the context of solving differential equations:
For a matrix \( A \), an eigenvector \( \mathbf{v} \) satisfies the equation \( A\mathbf{v} = \lambda\mathbf{v} \). Here, \( \lambda \) is the eigenvalue corresponding to this eigenvector.
In the context of solving differential equations:
- **Eigenvalues** determine the growth rates of the exponential terms in the solution. In our problem, the eigenvalues are \( 1, 2, \) and \( 5 \).
- **Eigenvectors** determine the direction of those exponential growths. They are essential for constructing the homogeneous solution.