Problem 30
Question
Use variation of parameters to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{ll}1 & -2 \\ 1 & -1\end{array}\right) \mathbf{x}+\left(\begin{array}{c}\tan t \\\ 1\end{array}\right)\)
Step-by-Step Solution
Verified Answer
Use variation of parameters to find particular solution, add to homogeneous solution.
1Step 1: Solve the Homogeneous Equation
First, solve the homogeneous system \( \mathbf{X}^{\prime} = \begin{pmatrix} 1 & -2 \ 1 & -1 \end{pmatrix} \mathbf{x} \). Find the eigenvalues by solving \( \text{det}(A - \lambda I) = 0 \), where \( A = \begin{pmatrix} 1 & -2 \ 1 & -1 \end{pmatrix} \). Calculating, the characteristic equation is \( \lambda^2 + \lambda = 0 \), giving eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = -1 \).
2Step 2: Find Eigenvectors
For \( \lambda_1 = 0 \), solve \( (A - 0 I) \mathbf{v} = 0 \). This gives \( \begin{pmatrix} 1 & -2 \ 1 & -1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \), resulting in \( \mathbf{v_1} = \begin{pmatrix} 2 \ 1 \end{pmatrix} \). For \( \lambda_2 = -1 \), solve \( (A + I) \mathbf{v} = 0 \), giving \( \mathbf{v_2} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
3Step 3: Construct the General Solution to the Homogeneous System
The general solution to the homogeneous system is \( \mathbf{x_h}(t) = c_1 \begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
4Step 4: Define the Particular Solution Using Variation of Parameters
Using variation of parameters, assume a solution of the form \( \mathbf{x_p}(t) = u_1(t) \begin{pmatrix} 2 \ 1 \end{pmatrix} + u_2(t) e^{-t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \). The derivatives must satisfy \( u_1'(t) \begin{pmatrix} 2 \ 1 \end{pmatrix} + u_2'(t) e^{-t} \begin{pmatrix} 1 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).
5Step 5: Solve for u'_1 and u'_2
Using the system of equations from the variation of parameters condition and substituting into the original non-homogeneous system, find \( u_1'(t) \) and \( u_2'(t) \) by solving \( \begin{pmatrix} 2 & e^{-t} \ 1 & e^{-t} \end{pmatrix} \begin{pmatrix} u_1'(t) \ u_2'(t) \end{pmatrix} = \begin{pmatrix} \tan t \ 1 \end{pmatrix} \).
6Step 6: Integrate to Find u_1 and u_2
Integrate \( u_1'(t) \) and \( u_2'(t) \) from the previous step to find \( u_1(t) \) and \( u_2(t) \).
7Step 7: Construct the Particular Solution and General Solution
Now plug \( u_1(t) \) and \( u_2(t) \) back into \( \mathbf{x_p}(t) = u_1(t) \begin{pmatrix} 2 \ 1 \end{pmatrix} + u_2(t) e^{-t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \) to get the particular solution. The general solution is then \( \mathbf{x}(t) = \mathbf{x_h}(t) + \mathbf{x_p}(t) \).
Key Concepts
Homogeneous EquationEigenvalues and EigenvectorsGeneral Solution
Homogeneous Equation
A homogeneous equation in the context of differential equations refers to a system that doesn't have any additional external forces or inputs. In simpler terms, it's like a set of equations that only depends on the variables themselves without any extra terms added. To solve such a system, we focus on finding the general behavior of the variables by examining their internal relationships.
In our exercise, we begin by considering the homogeneous equation given by \( \mathbf{X}^{\prime} = \begin{pmatrix} 1 & -2 \ 1 & -1 \end{pmatrix} \mathbf{x} \). This equations suggests a relationship solely defined by the matrix \( A \) and the variable \( \mathbf{x} \). Solving this involves finding the eigenvalues, which are special numbers that give clear insights into the behavior of the system.
The next step involves using these eigenvalues to find the eigenvectors. These vectors, when multiplied with the matrix, yield the same vectors scaled by the eigenvalue. These eigenvectors form the backbone of understanding the direction or flow of the system within its phase space.
In our exercise, we begin by considering the homogeneous equation given by \( \mathbf{X}^{\prime} = \begin{pmatrix} 1 & -2 \ 1 & -1 \end{pmatrix} \mathbf{x} \). This equations suggests a relationship solely defined by the matrix \( A \) and the variable \( \mathbf{x} \). Solving this involves finding the eigenvalues, which are special numbers that give clear insights into the behavior of the system.
The next step involves using these eigenvalues to find the eigenvectors. These vectors, when multiplied with the matrix, yield the same vectors scaled by the eigenvalue. These eigenvectors form the backbone of understanding the direction or flow of the system within its phase space.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are essential concepts in linear algebra that greatly simplify solving differential equations. They describe fundamental properties of matrices that help unfold the behavior of systems modeled by such matrices.
For the homogeneous equation, we calculate eigenvalues by resolving the characteristic equation \( \text{det}(A - \lambda I) = 0 \). Plugging our matrix \( A \) into this yields the characteristic polynomial \( \lambda^2 + \lambda = 0 \), granting us the eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = -1 \).
Once eigenvalues are found, the next task is finding the corresponding eigenvectors. For \( \lambda_1 = 0 \), solving \( (A - 0I) \mathbf{v_1} = 0 \) results in the eigenvector \( \mathbf{v_1} = \begin{pmatrix} 2 \ 1 \end{pmatrix} \). Similarly, for \( \lambda_2 = -1 \), the process gives \( \mathbf{v_2} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
These pairs of eigenvalues and eigenvectors hold immense significance as they describe how solutions evolve within the system over time. They also simplify a system of equations into manageable parts, highlighting primary directions of influence in multi-variable environments.
For the homogeneous equation, we calculate eigenvalues by resolving the characteristic equation \( \text{det}(A - \lambda I) = 0 \). Plugging our matrix \( A \) into this yields the characteristic polynomial \( \lambda^2 + \lambda = 0 \), granting us the eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = -1 \).
Once eigenvalues are found, the next task is finding the corresponding eigenvectors. For \( \lambda_1 = 0 \), solving \( (A - 0I) \mathbf{v_1} = 0 \) results in the eigenvector \( \mathbf{v_1} = \begin{pmatrix} 2 \ 1 \end{pmatrix} \). Similarly, for \( \lambda_2 = -1 \), the process gives \( \mathbf{v_2} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
These pairs of eigenvalues and eigenvectors hold immense significance as they describe how solutions evolve within the system over time. They also simplify a system of equations into manageable parts, highlighting primary directions of influence in multi-variable environments.
General Solution
The general solution of a differential system aims to capture all possible scenarios of its behavior by combining solutions derived from both homogeneous and non-homogeneous parts of the equation.
For our homogeneous system, the general solution is expressed as \( \mathbf{x_h}(t) = c_1 \begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \). These terms represent a combination of the basis solutions tailored depending on the initial conditions defined by \( c_1 \) and \( c_2 \).
To extend this solution to a non-homogeneous system, a particular solution must be constructed, typically using variation of parameters. This adds a level of customization specific to the external factors affecting the system, such as the additional vector \( \begin{pmatrix} \tan t \ 1 \end{pmatrix} \). After determining these particular parameters, integrate them into the general solution as shown: \( \mathbf{x}(t) = \mathbf{x_h}(t) + \mathbf{x_p}(t) \).
Thus, the complete general solution embodies both the natural behavior described by \( \mathbf{x_h}(t) \) and adjustments for influences captured by \( \mathbf{x_p}(t) \), rendering a complete picture of the system's dynamics.
For our homogeneous system, the general solution is expressed as \( \mathbf{x_h}(t) = c_1 \begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \). These terms represent a combination of the basis solutions tailored depending on the initial conditions defined by \( c_1 \) and \( c_2 \).
To extend this solution to a non-homogeneous system, a particular solution must be constructed, typically using variation of parameters. This adds a level of customization specific to the external factors affecting the system, such as the additional vector \( \begin{pmatrix} \tan t \ 1 \end{pmatrix} \). After determining these particular parameters, integrate them into the general solution as shown: \( \mathbf{x}(t) = \mathbf{x_h}(t) + \mathbf{x_p}(t) \).
Thus, the complete general solution embodies both the natural behavior described by \( \mathbf{x_h}(t) \) and adjustments for influences captured by \( \mathbf{x_p}(t) \), rendering a complete picture of the system's dynamics.
Other exercises in this chapter
Problem 29
In Problems 13-32, use vaniation of parameters to solve the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} 1 & 2 \\ -\frac{1}{2} & 1 \end{array}\r
View solution Problem 29
In Problems 21-30, find the general solution of the given system. $$ X^{\prime}=\left(\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 2 & -1 \\ 0 & 1 & 0 \end{array}\right)
View solution Problem 30
Find the general solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{lll} 4 & 1 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{array}\right) \mathbf{X}
View solution Problem 30
In Problems 13-32, use vaniation of parameters to solve the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{ll} 1 & -2 \\ 1 & -1 \end{array}\right) \ma
View solution