Chapter 11: Problem 24
Solve the given initial-value problem. $$ \left[\begin{array}{l} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} -3 & 4 \\ -1 & 2 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=1 \text { and } x_{2}(0)=2 \text { . } $$
Short Answer
Step by step solution
Write the System as a Matrix Equation
Solve for Eigenvalues of the Matrix
Find the Eigenvectors
Form the General Solution
Apply Initial Conditions
Substitute Values to Find Particular Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues
In step 2 of our problem, we computed the eigenvalues \( \lambda \) for the matrix \( A \). The characteristic equation is derived from the determinant: - First, subtract \( \lambda \) from the diagonal elements of the matrix. - Then, compute the determinant, which involves multiplying the diagonal elements and subtracting the product of the off-diagonal elements.
For the matrix \( A \) given, this results in the quadratic equation \( \lambda^2 + \lambda - 10 = 0 \). Solving this equation gave us the eigenvalues \( \lambda = 2 \) and \( \lambda = -5 \), each of which describes a different dynamic behavior of the system.
Eigenvectors
This equation helps identify the vectors that remain in the same span (line through the origin) under the transformation described by \( A \). Practically, this means setting up a system of linear equations using the found eigenvalue and solving for vector \( \mathbf{v} \).
For our exercise, once \( \lambda \) values were identified:
- We found the first eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 4 \ 1 \end{bmatrix} \) for \( \lambda = 2 \)
- And the second eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 2 \ -1 \end{bmatrix} \) for \( \lambda = -5 \)
Matrix Equations
The system of differential equations we began with is represented in matrix form as \( \mathbf{x}'(t) = A\mathbf{x}(t) \). Here, \( \mathbf{x}(t) \) is a vector of functions, and \( A \) is a square matrix. This setup translates the problem of finding solutions to the differential equation into a problem of linear algebra.
The matrix equation setup helps:
- Identify eigenvalues and eigenvectors which provide the solution's structure
- Simplify the process of handling simultaneous differential equations