Problem 17
Question
Find the general solution of each given system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line the direction in which the solution would move if it starts on that line. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} -2 & 0 \\ -3 & 1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The general solution is \( \mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \).
1Step 1: Identify the Matrix
The system of differential equations can be expressed in matrix form as \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), where \( A = \begin{bmatrix} -2 & 0 \ -3 & 1 \end{bmatrix} \). The vector \( \mathbf{x}(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) represents the variables of the system.
2Step 2: Find the Eigenvalues of the Matrix
The eigenvalues \( \lambda \) of the matrix \( A \) satisfy the characteristic equation \( \det(A - \lambda I) = 0 \). For the given matrix, this becomes: \[ \det \begin{bmatrix} -2 - \lambda & 0 \ -3 & 1 - \lambda \end{bmatrix} = (\lambda + 2)(\lambda - 1) = 0 \] Solving this gives the eigenvalues \( \lambda_1 = -2 \) and \( \lambda_2 = 1 \).
3Step 3: Find the Eigenvectors for Each Eigenvalue
For each eigenvalue \( \lambda_i \), find the corresponding eigenvector \( \mathbf{v}_i \) by solving \( (A - \lambda_i I)\mathbf{v}_i = \mathbf{0} \). - For \( \lambda_1 = -2 \): Solve \( \begin{bmatrix} 0 & 0 \ -3 & 3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This gives \( v_1 = 0 \) and \( v_2 \) is free, so an eigenvector is \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \).- For \( \lambda_2 = 1 \): Solve \( \begin{bmatrix} -3 & 0 \ -3 & 0 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This gives \( v_2 = 0 \) and \( v_1 \) is free, so an eigenvector is \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \).
4Step 4: Form the General Solution
The general solution of the system can be written as a linear combination of the eigenvectors, each multiplied by an exponential function of their respective eigenvalue:\( \mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 0 \ 1 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 1 \ 0 \end{bmatrix} \), where \( c_1 \) and \( c_2 \) are arbitrary constants.
5Step 5: Sketch Direction Lines and Their Solutions
Eigenvector \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \) corresponds to solutions moving along the \( x_2 \) axis, decreasing since \( e^{-2t} \) decreases over time. Eigenvector \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) corresponds to solutions moving along the \( x_1 \) axis, increasing since \( e^{t} \) increases over time. Draw these lines in the direction of their respective eigenvectors and indicate the direction of increasing or decreasing motion along these lines.
Key Concepts
EigenvaluesEigenvectorsGeneral Solution
Eigenvalues
Eigenvalues are fundamental in understanding systems of differential equations. They provide insight into how a system behaves over time. For a matrix, eigenvalues are the special numbers that reveal information about the transformation it represents. When solving differential equations in the form \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), discovering the eigenvalues of matrix \( A \) becomes crucial.
The characteristic equation \( \det(A - \lambda I) = 0 \) helps us find eigenvalues. Here, \( I \) is the identity matrix, and \( \lambda \) represents an eigenvalue. For example, for a matrix \( A = \begin{bmatrix} -2 & 0 \ -3 & 1 \end{bmatrix} \), solving the characteristic equation \( (\lambda + 2)(\lambda - 1) = 0 \) gives the eigenvalues \( \lambda_1 = -2 \) and \( \lambda_2 = 1 \).
Understanding eigenvalues helps describe how each component of a solution grows or decays. In this case, the eigenvalue \( -2 \) suggests a decaying behavior, while \( 1 \) leads to growth. This foundational step sets the stage for further analysis and interpretation of the system's dynamics.
The characteristic equation \( \det(A - \lambda I) = 0 \) helps us find eigenvalues. Here, \( I \) is the identity matrix, and \( \lambda \) represents an eigenvalue. For example, for a matrix \( A = \begin{bmatrix} -2 & 0 \ -3 & 1 \end{bmatrix} \), solving the characteristic equation \( (\lambda + 2)(\lambda - 1) = 0 \) gives the eigenvalues \( \lambda_1 = -2 \) and \( \lambda_2 = 1 \).
Understanding eigenvalues helps describe how each component of a solution grows or decays. In this case, the eigenvalue \( -2 \) suggests a decaying behavior, while \( 1 \) leads to growth. This foundational step sets the stage for further analysis and interpretation of the system's dynamics.
Eigenvectors
Eigenvectors accompany eigenvalues and further aid in understanding the behavior of differential equations. Once we have the eigenvalues, we need to find the eigenvectors. These special vectors do not change direction when the matrix transformation is applied, only potentially their magnitude.
For a given eigenvalue \( \lambda_i \), the eigenvector is found using the equation \( (A - \lambda_i I)\mathbf{v}_i = \mathbf{0} \). In simple terms, we solve this equation to find vectors that show how solutions to the differential equations behave. For our matrix \( A \), we have two eigenvectors:
For a given eigenvalue \( \lambda_i \), the eigenvector is found using the equation \( (A - \lambda_i I)\mathbf{v}_i = \mathbf{0} \). In simple terms, we solve this equation to find vectors that show how solutions to the differential equations behave. For our matrix \( A \), we have two eigenvectors:
- For \( \lambda_1 = -2 \), the eigenvector is \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \), which suggests a transformation along the \( x_2 \) axis.
- For \( \lambda_2 = 1 \), the eigenvector is \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \), indicating changes along the \( x_1 \) axis.
General Solution
The general solution of a differential equation system describes all possible behaviors of the system over time. It is constructed by combining the eigenvectors with their respective exponential growth or decay functions derived from the eigenvalues. This combination is represented as:
\[ \mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 0 \ 1 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 1 \ 0 \end{bmatrix} \]
Here, \( c_1 \) and \( c_2 \) are arbitrary constants representing initial conditions. This is important because any specific initial state can be expressed with a particular set of constants.
The exponential terms \( e^{-2t} \) and \( e^{t} \) indicate how each component of the solution changes over time. The term \( e^{-2t} \) implies decay, while \( e^{t} \) suggests growth. By graphing these solutions, you can visualize how they move and change, offering a clear picture of the system's dynamic behavior. Understanding the general solution is vital in predicting how the system's state evolves with time, especially when visualizing stability and direction.
\[ \mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 0 \ 1 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 1 \ 0 \end{bmatrix} \]
Here, \( c_1 \) and \( c_2 \) are arbitrary constants representing initial conditions. This is important because any specific initial state can be expressed with a particular set of constants.
The exponential terms \( e^{-2t} \) and \( e^{t} \) indicate how each component of the solution changes over time. The term \( e^{-2t} \) implies decay, while \( e^{t} \) suggests growth. By graphing these solutions, you can visualize how they move and change, offering a clear picture of the system's dynamic behavior. Understanding the general solution is vital in predicting how the system's state evolves with time, especially when visualizing stability and direction.
Other exercises in this chapter
Problem 16
Find the general solution of each given system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line the di
View solution Problem 16
Assume that $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(10-x_{1}-2 x_{2}\right) \\ \frac{d x_{2}}{d t}=x_{2}\left(10-2 x_{1}-x_{2}\right) \end{array} $$
View solution Problem 17
Assume that \(N(t)\) denotes the density of an insect species at time \(t\) and \(P(t)\) denotes the density of its predator at time \(t\). The insect species i
View solution Problem 17
$$ \text { Systems of Differential Equations } $$ $$ \begin{array}{l} \frac{d x_{1}}{d t}=-0.2 x_{1} \\ \frac{d x_{2}}{d t}=-0.3 x_{2} \end{array} $$
View solution