Problem 1
Question
The general solution of the linear system \(\mathbf{X}^{\prime}=\mathbf{A} \mathbf{X}\) is given. (a) In each case discuss the nature of the solution in a neighborhood of \((0,0)\). (b) With the aid of a graphing utility plot the solution that satisfies \(\mathbf{X}(0)=(1,1)\) $$ \mathbf{A}=\left(\begin{array}{ll} -2 & -2 \\ -2 & -5 \end{array}\right), \quad \mathbf{X}(t)=c_{1}\left(\begin{array}{r} 2 \\ -1 \end{array}\right) e^{-t}+c_{2}\left(\begin{array}{l} 1 \\ 2 \end{array}\right) e^{-6 t} $$
Step-by-Step Solution
Verified Answer
The eigenvalues are negative, indicating a stable node at the origin. The particular solution approaches the origin as time increases.
1Step 1: Eigenvalues and Eigenvectors
First, find the eigenvalues of the matrix \( \mathbf{A} = \begin{pmatrix} -2 & -2 \ -2 & -5 \end{pmatrix} \). To find the eigenvalues, solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This results in the quadratic equation \( (-2-\lambda)(-5-\lambda) - 4 = 0 \), which simplifies to \( \lambda^2 + 7\lambda + 6 = 0 \). Solving this gives two eigenvalues: \( \lambda_1 = -1 \) and \( \lambda_2 = -6 \).
2Step 2: Examine the Nature of Solution
The solution of the system is related to its eigenvalues. Here, we have two real, distinct eigenvalues: \( \lambda_1 = -1 \) and \( \lambda_2 = -6 \), both negative. This indicates that the origin \((0,0)\) is a stable node. Solutions will tend towards the origin over time, as both eigenvalues are negative.
3Step 3: Particular Solution for Given Initial Condition
To satisfy \( \mathbf{X}(0) = (1,1) \), use the provided general solution:\[ \mathbf{X}(t) = c_1 \begin{pmatrix} 2 \ -1 \end{pmatrix} e^{-t} + c_2 \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{-6t} \]At \( t = 0 \), \( \mathbf{X}(0) = c_1 \begin{pmatrix} 2 \ -1 \end{pmatrix} + c_2 \begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).This results in two equations: - \( 2c_1 + c_2 = 1 \)- \( -c_1 + 2c_2 = 1 \)Solving these equations yields \( c_1 = \frac{1}{5} \), and \( c_2 = \frac{3}{5} \).
4Step 4: Plot the Solution Using a Graphing Utility
Now that we have the constants \( c_1 = \frac{1}{5} \) and \( c_2 = \frac{3}{5} \), the particular solution is \[ \mathbf{X}(t) = \frac{1}{5} \begin{pmatrix} 2 \ -1 \end{pmatrix} e^{-t} + \frac{3}{5} \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{-6t} \]Use a graphing utility to plot this solution, starting at the point \((1,1)\) at \( t=0 \), and observe how it approaches the origin as \( t \) increases.
Key Concepts
Eigenvalues and EigenvectorsStable NodeCharacteristic EquationSolution of Linear Systems
Eigenvalues and Eigenvectors
In the realm of linear differential equations, understanding eigenvalues and eigenvectors is crucial. When you have a system expressed as \( \mathbf{X}^{\prime} = \mathbf{A} \mathbf{X} \), where \( \mathbf{A} \) is a matrix, the first thing to do is to look for its eigenvalues and eigenvectors.
To find the eigenvalues, you need to solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Here, \( \lambda \) represents eigenvalues, and \( \mathbf{I} \) is the identity matrix of the same dimensions as \( \mathbf{A} \). The determinant calculated will yield a polynomial equation, whose roots are the eigenvalues of the matrix.
Once you have the eigenvalues, eigenvectors are found by solving the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \), where \( \mathbf{v} \) is the eigenvector corresponding to the eigenvalue \( \lambda \). This involves substituting the eigenvalues back into the equation and solving the resulting set of linear equations.
To find the eigenvalues, you need to solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Here, \( \lambda \) represents eigenvalues, and \( \mathbf{I} \) is the identity matrix of the same dimensions as \( \mathbf{A} \). The determinant calculated will yield a polynomial equation, whose roots are the eigenvalues of the matrix.
Once you have the eigenvalues, eigenvectors are found by solving the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \), where \( \mathbf{v} \) is the eigenvector corresponding to the eigenvalue \( \lambda \). This involves substituting the eigenvalues back into the equation and solving the resulting set of linear equations.
Stable Node
The concept of a stable node arises when analyzing the behavior of solutions near an equilibrium point, such as \((0,0)\). It provides insight into the stability of solutions over time.
In our scenario, both eigenvalues were real and negative \((\lambda_1 = -1, \lambda_2 = -6)\). This specific set of eigenvalue characteristics indicates that the origin is a stable node. It means that trajectories or solutions near the origin will be drawn towards it as time progresses.
A stable node provides a dampening effect, where any small perturbation in the system will naturally decrease, returning to the equilibrium point. Key indicators of a stable node include:
In our scenario, both eigenvalues were real and negative \((\lambda_1 = -1, \lambda_2 = -6)\). This specific set of eigenvalue characteristics indicates that the origin is a stable node. It means that trajectories or solutions near the origin will be drawn towards it as time progresses.
A stable node provides a dampening effect, where any small perturbation in the system will naturally decrease, returning to the equilibrium point. Key indicators of a stable node include:
- All eigenvalues must be real and negative.
- Solutions converge to a single point over time.
- The system's dynamics are inherently self-stabilizing.
Characteristic Equation
The characteristic equation is a pivotal part of solving linear systems and finding eigenvalues. It forms the bridge between the matrix \( \mathbf{A} \) of a differential equation and the solution properties of the system.
To derive the characteristic equation, you calculate the determinant \( \det(\mathbf{A} - \lambda \mathbf{I}) \) and set it to zero. In our case, this calculation resulted in the quadratic equation:\[\lambda^2 + 7\lambda + 6 = 0\]This polynomial reflects the dynamics encapsulated by the matrix \( \mathbf{A} \) and is fundamental in determining the eigenvalues.
Once the polynomial is derived, solving it will provide the eigenvalues \( \lambda \). These eigenvalues determine the nature of the solutions:
To derive the characteristic equation, you calculate the determinant \( \det(\mathbf{A} - \lambda \mathbf{I}) \) and set it to zero. In our case, this calculation resulted in the quadratic equation:\[\lambda^2 + 7\lambda + 6 = 0\]This polynomial reflects the dynamics encapsulated by the matrix \( \mathbf{A} \) and is fundamental in determining the eigenvalues.
Once the polynomial is derived, solving it will provide the eigenvalues \( \lambda \). These eigenvalues determine the nature of the solutions:
- If all eigenvalues are real and negative, the system tends to be stable.
- Complex eigenvalues introduce oscillatory behavior.
- Zero or positive real parts of eigenvalues often indicate instability.
Solution of Linear Systems
In linear algebra and differential equations, solving a system involves expressing the state of a system over time using initial conditions and the system’s defining parameters.
For the system \( \mathbf{X}^{\prime} = \mathbf{A} \mathbf{X} \), once eigenvalues and eigenvectors are known, you can construct the general solution. Our system's general solution is given by:\[\mathbf{X}(t) = c_1 \begin{pmatrix} 2 \ -1 \end{pmatrix} e^{-t} + c_2 \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{-6t}\]Where \( c_1 \) and \( c_2 \) are constants determined by initial conditions. These constants tailor the general solution to meet specific criteria at \( t = 0 \), like \((1, 1)\) in this case.
Systems modeled this way can represent diverse scenarios, from mechanical oscillations to electrical circuits or population dynamics in ecology. To find the specific solution that fits your initial conditions, solve:
For the system \( \mathbf{X}^{\prime} = \mathbf{A} \mathbf{X} \), once eigenvalues and eigenvectors are known, you can construct the general solution. Our system's general solution is given by:\[\mathbf{X}(t) = c_1 \begin{pmatrix} 2 \ -1 \end{pmatrix} e^{-t} + c_2 \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{-6t}\]Where \( c_1 \) and \( c_2 \) are constants determined by initial conditions. These constants tailor the general solution to meet specific criteria at \( t = 0 \), like \((1, 1)\) in this case.
Systems modeled this way can represent diverse scenarios, from mechanical oscillations to electrical circuits or population dynamics in ecology. To find the specific solution that fits your initial conditions, solve:
- Set initial conditions \( \mathbf{X}(0) = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
- Substitute to get equations for \( c_1 \) and \( c_2 \).
- Solve this algebraically to apply these constants into your general formula.
Other exercises in this chapter
Problem 1
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