Chapter 11: Problem 23
Solve the given initial-value problem. \(\left[\begin{array}{l}\frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t}\end{array}\right]=\left[\begin{array}{cc}4 & -7 \\ 2 & -5\end{array}\right]\left[\begin{array}{l}x_{1}(t) \\\ x_{2}(t)\end{array}\right]\) with \(x_{1}(0)=13\) and \(x_{2}(0)=3\).
Short Answer
Expert verified
The solution is \( x(t) = 2e^{2t}\begin{bmatrix} 7 \\ 2 \end{bmatrix} - e^{-3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix} \)."
Step by step solution
01
Write the System of Equations
The system of differential equations is given as \(\frac{d x_1}{dt} = 4x_1 - 7x_2\) and \(\frac{d x_2}{dt} = 2x_1 - 5x_2\). This matrix form involves the vector \(\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\) and the matrix \(\begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix}\).
02
Find the Eigenvalues of the Coefficient Matrix
Calculate the eigenvalues of the matrix \(\begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix}\). The eigenvalues can be found by solving the characteristic equation: \(\det(\begin{bmatrix} 4-\lambda & -7 \ 2 & -5-\lambda \end{bmatrix}) = 0\).
03
Solve the Characteristic Equation
The characteristic equation is \((4 - \lambda)(-5 - \lambda) - (-7)(2) = 0\). Simplifying yields \(\lambda^2 + \lambda - 6 = 0\). Factor the quadratic to get \( (\lambda - 2)(\lambda + 3) = 0 \), giving the eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = -3\).
04
Find the Eigenvectors
For each eigenvalue, solve \((A - \lambda I)\mathbf{v} = \mathbf{0}\). For \(\lambda_1 = 2\), solve \(\begin{bmatrix} 2 & -7 \ 2 & -7 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), resulting in eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 7 \ 2 \end{bmatrix}\). For \(\lambda_2 = -3\), solve \(\begin{bmatrix} 7 & -7 \ 2 & -2 \end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), resulting in eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\).
05
Form the General Solution
The general solution is \( \mathbf{x}(t) = c_1 e^{2t} \begin{bmatrix} 7 \ 2 \end{bmatrix} + c_2 e^{-3t} \begin{bmatrix} 1 \ 1 \end{bmatrix} \). This combines the eigenvectors with exponential functions of eigenvalues.
06
Apply Initial Conditions
Use initial conditions \(x_1(0)=13\) and \(x_2(0)=3\) to solve for constants \(c_1\) and \(c_2\). At \(t = 0\), the system becomes \(13 = 7c_1 + c_2\) and \(3 = 2c_1 + c_2\). Solve the system to get \(c_1 = 2\) and \(c_2 = -1\).
07
Write the Particular Solution
Substitute the constants \(c_1 = 2\) and \(c_2 = -1\) back into the general solution to get the specific solution: \( \mathbf{x}(t) = 2e^{2t} \begin{bmatrix} 7 \ 2 \end{bmatrix} - e^{-3t} \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Differential Equations
In the context of mathematics, a system of differential equations is a collection of two or more interrelated differential equations. These equations describe how multiple functions change over time, interacting with each other. Consider our example where we have:
The beauty of using matrices lies in their power to summarize complex systems succinctly. The initial value problem, with its condition specifying values at \( t = 0 \), further guides us to specific solutions from the general solutions possible.
- \( \frac{d x_1}{dt} = 4x_1 - 7x_2 \)
- \( \frac{d x_2}{dt} = 2x_1 - 5x_2 \)
The beauty of using matrices lies in their power to summarize complex systems succinctly. The initial value problem, with its condition specifying values at \( t = 0 \), further guides us to specific solutions from the general solutions possible.
Eigenvalues and Eigenvectors
To solve a system of differential equations, finding eigenvalues and eigenvectors is a crucial step. Eigenvalues indicate the specific scaling factors of transformations, whereas eigenvectors determine the direction of these transformations.
In our example, we consider the matrix:
For these eigenvalues, we identify the eigenvectors:
In our example, we consider the matrix:
- \( \begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix} \)
- Eigenvalues: \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \)
For these eigenvalues, we identify the eigenvectors:
- For \( \lambda_1 = 2 \), the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 7 \ 2 \end{bmatrix} \)
- For \( \lambda_2 = -3 \), the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \)
Characteristic Equation
The characteristic equation is pivotal in finding eigenvalues of a matrix, and hence, essential in solving systems of differential equations. It is derived from the determinant of the matrix \( A - \lambda I \), where \( I \) denotes the identity matrix.
For our system, the matrix:
Thus, the characteristic equation not only leads us to the eigenvalues but ultimately unveils important dynamics of the system's behavior.
For our system, the matrix:
- \( \begin{bmatrix} 4 & -7 \ 2 & -5 \end{bmatrix} \)
Thus, the characteristic equation not only leads us to the eigenvalues but ultimately unveils important dynamics of the system's behavior.