Problem 13
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. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 1 & 3 \\ 5 & 3 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$
Step-by-Step Solution
Verified Answer
Eigenvalues: \( \lambda_1 = 6, \lambda_2 = -2 \). General solution: \( \mathbf{x}(t) = c_1 \begin{bmatrix} 3 \\ 5 \end{bmatrix} e^{6t} + c_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix} e^{-2t} \).
1Step 1: Write the System in Matrix Form
The system of differential equations is given as: \[ \frac{d}{dt} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 1 & 3 \ 5 & 3 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \]. This representation helps in understanding how the states \(x_1\) and \(x_2\) evolve over time.
2Step 2: Find the Eigenvalues
To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( A = \begin{bmatrix} 1 & 3 \ 5 & 3 \end{bmatrix} \). Calculate \((A - \lambda I)\): \[ \begin{vmatrix} 1-\lambda & 3 \ 5 & 3-\lambda \end{vmatrix} = (1-\lambda)(3-\lambda) - 3 \cdot 5 \]. This simplifies to \( \lambda^2 - 4\lambda - 12 = 0 \). Then, solve for \( \lambda \): \( \lambda_1 = 6, \lambda_2 = -2 \).
3Step 3: Find the Eigenvectors
For each eigenvalue, find the corresponding eigenvector \(v\) by solving \((A - \lambda I)v = 0\). For \( \lambda_1 = 6 \), solve: \[ \begin{bmatrix} -5 & 3 \ 5 & -3 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = 0 \]. The solution gives the eigenvector \( v_1 = \begin{bmatrix} 3 \ 5 \end{bmatrix} \). For \( \lambda_2 = -2 \), solve: \[ \begin{bmatrix} 3 & 3 \ 5 & 5 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = 0 \]. The solution gives the eigenvector \( v_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix}\).
4Step 4: Write the General Solution
The solution of the system can be written using the eigenvectors and eigenvalues as: \[ \mathbf{x}(t) = c_1 \begin{bmatrix} 3 \ 5 \end{bmatrix} e^{6t} + c_2 \begin{bmatrix} -1 \ 1 \end{bmatrix} e^{-2t} \], where \(c_1\) and \(c_2\) are constants determined by initial conditions.
5Step 5: Sketch the Solution and Indicate Directions
Draw the eigenvectors and sketch lines along them on the x1-x2 plane. For \( v_1 = \begin{bmatrix} 3 \ 5 \end{bmatrix} \), the line moves quickly outwards (positive matrix exponent). For \( v_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix} \), the solutions decay (negative matrix exponent). Indicate arrows showing increasing and decreasing direction based on their exponentials.
Key Concepts
EigenvaluesEigenvectorsCharacteristic Equation
Eigenvalues
When dealing with systems of differential equations, eigenvalues play a crucial role in understanding the behavior of the solutions. An eigenvalue is a special number associated with a matrix that provides insights into the system's properties. For a given matrix, eigenvalues are found by solving the characteristic equation. This process involves finding the values of \( \lambda \) for which the determinant of \( A - \lambda I \) is zero, where \( A \) is the matrix derived from the system, and \( I \) is the identity matrix of the same dimension.
In the problem at hand, we have the matrix \[ A = \begin{bmatrix} 1 & 3 \ 5 & 3 \end{bmatrix} \]. By setting up the characteristic equation \( \det(A - \lambda I) = 0 \), we find the eigenvalues by solving \( (1-\lambda)(3-\lambda) - 15 = 0 \). This simplification leads us to a quadratic equation \( \lambda^2 - 4\lambda - 12 = 0 \). The solutions to this equation are the eigenvalues \( \lambda_1 = 6 \) and \( \lambda_2 = -2 \).
These eigenvalues indicate the rate of change in the direction of their corresponding eigenvectors. Specifically, a positive eigenvalue suggests growth, while a negative one implies decay.
In the problem at hand, we have the matrix \[ A = \begin{bmatrix} 1 & 3 \ 5 & 3 \end{bmatrix} \]. By setting up the characteristic equation \( \det(A - \lambda I) = 0 \), we find the eigenvalues by solving \( (1-\lambda)(3-\lambda) - 15 = 0 \). This simplification leads us to a quadratic equation \( \lambda^2 - 4\lambda - 12 = 0 \). The solutions to this equation are the eigenvalues \( \lambda_1 = 6 \) and \( \lambda_2 = -2 \).
These eigenvalues indicate the rate of change in the direction of their corresponding eigenvectors. Specifically, a positive eigenvalue suggests growth, while a negative one implies decay.
Eigenvectors
Once eigenvalues are determined, the next step is to find the corresponding eigenvectors. Eigenvectors are vectors that describe the direction in which the transformation described by the matrix acts as merely a scaling operation. Their significance within systems of differential equations is in indicating the directions along which the solutions predominantly move.
For each eigenvalue \( \lambda \), eigenvectors are found by solving the equation \( (A - \lambda I)v = 0 \). For our specific system, we have two eigenvalues: 6 and -2. Let's consider \( \lambda_1 = 6 \). We construct the matrix \( A - 6I = \begin{bmatrix} -5 & 3 \ 5 & -3 \end{bmatrix} \) and solve the linear system to find the eigenvector, resulting in \( v_1 = \begin{bmatrix} 3 \ 5 \end{bmatrix} \).
Similarly, for \( \lambda_2 = -2 \), we form \( A + 2I = \begin{bmatrix} 3 & 3 \ 5 & 5 \end{bmatrix} \), solving which gives the eigenvector \( v_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix} \). These eigenvectors indicate the paths along the component of solution vector \( \mathbf{x}(t) \) tending to grow or decay.
For each eigenvalue \( \lambda \), eigenvectors are found by solving the equation \( (A - \lambda I)v = 0 \). For our specific system, we have two eigenvalues: 6 and -2. Let's consider \( \lambda_1 = 6 \). We construct the matrix \( A - 6I = \begin{bmatrix} -5 & 3 \ 5 & -3 \end{bmatrix} \) and solve the linear system to find the eigenvector, resulting in \( v_1 = \begin{bmatrix} 3 \ 5 \end{bmatrix} \).
Similarly, for \( \lambda_2 = -2 \), we form \( A + 2I = \begin{bmatrix} 3 & 3 \ 5 & 5 \end{bmatrix} \), solving which gives the eigenvector \( v_2 = \begin{bmatrix} -1 \ 1 \end{bmatrix} \). These eigenvectors indicate the paths along the component of solution vector \( \mathbf{x}(t) \) tending to grow or decay.
Characteristic Equation
The characteristic equation is a mathematical expression derived as part of the process to find eigenvalues. In the context of systems of differential equations, it provides a methodical way to ascertain the stability and behavior of the system by identifying eigenvalues.
To form the characteristic equation, we take the given system's matrix \( A \) and subtract \( \lambda \) times the identity matrix \( I \). The equation reads as \( \det(A - \lambda I) = 0 \), where the determinant calculates a special scalar value summarizing the matrix's properties related to its linear transformations.
For the exercise, \( A = \begin{bmatrix} 1 & 3 \ 5 & 3 \end{bmatrix} \) means calculating \( \det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 3 \ 5 & 3-\lambda \end{vmatrix} \). Expanding this determinant yields \[ (1-\lambda)(3-\lambda) - 15 = \lambda^2 - 4\lambda - 12 \], the characteristic equation. Solving this quadratic equation finds the eigenvalues; in our case, \( \lambda_1 = 6 \) and \( \lambda_2 = -2 \). These solutions help determine the type of system behavior, enabling further analysis and solution sketching.
To form the characteristic equation, we take the given system's matrix \( A \) and subtract \( \lambda \) times the identity matrix \( I \). The equation reads as \( \det(A - \lambda I) = 0 \), where the determinant calculates a special scalar value summarizing the matrix's properties related to its linear transformations.
For the exercise, \( A = \begin{bmatrix} 1 & 3 \ 5 & 3 \end{bmatrix} \) means calculating \( \det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 3 \ 5 & 3-\lambda \end{vmatrix} \). Expanding this determinant yields \[ (1-\lambda)(3-\lambda) - 15 = \lambda^2 - 4\lambda - 12 \], the characteristic equation. Solving this quadratic equation finds the eigenvalues; in our case, \( \lambda_1 = 6 \) and \( \lambda_2 = -2 \). These solutions help determine the type of system behavior, enabling further analysis and solution sketching.
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