Chapter 5: Problem 12
Find general solutions of the systems in Problems 1 through 22\. In Problems I through 6, use a computer system or graphing calculator to construct a direction field and typical solution curves for the given system. \(\mathbf{x}^{\prime}=\left[\begin{array}{rrr}-1 & 0 & 1 \\ 0 & -1 & 1 \\ 1 & -1 & -1\end{array}\right] \mathbf{x}\)
Short Answer
Step by step solution
Understand the Matrix Representation
Find the Eigenvalues of A
Solve the Characteristic Equation
Simplify and Solve for Eigenvalues
Find Eigenvectors Corresponding to Eigenvalues
Form the General Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Representation
Consider a vector \(\mathbf{x} = \left[x_1(t), x_2(t), x_3(t)\right]^T\). Each function \(x_i(t)\) represents a differential equation in the system. The matrix \(A\) contains the coefficients that multiply each of these functions and their derivatives.
- In our example, the matrix \(A\) is \(\begin{bmatrix} -1 & 0 & 1 \0 & -1 & 1 \1 & -1 & -1 \end{bmatrix}\).
- Each row in \(A\) affects one of the systems of equations, combining influences from each of the \(x_1(t)\), \(x_2(t)\), and \(x_3(t)\) terms.
Eigenvalues and Eigenvectors
To find eigenvalues, we solve the characteristic equation \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues.
- In our example, solving this equation results in the eigenvalues: \(\lambda_1 = 0\), \(\lambda_2 = \frac{-3 + \sqrt{3}i}{2}\), and \(\lambda_3 = \frac{-3 - \sqrt{3}i}{2}\).
- The solutions include both real and complex values indicating different dynamics like oscillations or growth/decay trends based on the system state.
- Eigenvectors act as direction indicators, showing the path that solutions will trend towards over time.
Characteristic Equation
The process involves calculating the determinant of \(A - \lambda I\) and setting it equal to zero: \[ |A - \lambda I| = 0 \]. By doing this, we derive a polynomial whose roots are the eigenvalues of \(A\).
- For our system, the matrix \(A\) yields the characteristic polynomial \(\lambda^3 + 3\lambda^2 + 3\lambda = 0\).
- This factors to \(\lambda(\lambda^2 + 3\lambda + 3) = 0\), leading us to determine the eigenvalues.
General Solution of Differential Equations
For our system, the general solution takes the form: \[ \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 + c_3 e^{\lambda_3 t} \mathbf{v}_3 \], where each term \(c_i e^{\lambda_i t} \mathbf{v}_i\) represents a scaled solution path.
- The constants \(c_i\) depend on the initial conditions, dictating how much influence each path has.
- In cases with complex eigenvalues, expect terms involving oscillatory elements, like sines and cosines.