Chapter 7: Problem 2
Express the general solution of the given system of equations in terms of real-valued functions. In each of Problems 1 through 6 also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-1} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x} $$
Short Answer
Expert verified
Solving the characteristic equation, we get:
$$
(-1-\lambda)((-1)-\lambda) - (-4)(1) = 0
$$
Expanding and simplifying,
$$
\lambda^2 + 2\lambda + 1 - 4 = 0 \Rightarrow \lambda^2 + 2\lambda - 3 = 0
$$
Factoring the quadratic equation gives:
$$
(\lambda - 1)(\lambda + 3) = 0
$$
So, the eigenvalues are:
$$
\lambda_1 = 1 \quad \text{and} \quad \lambda_2 = -3
$$
#tag_title#Step 2: Calculate the eigenvectors of the matrix#tag_content#Now we'll find the eigenvectors for each eigenvalue. For each eigenvalue, we'll solve the following system of linear equations:
$$
(A - \lambda I)\vec{v} = 0
$$
For $\lambda_1 = 1,$ we have:
$$
\left(\begin{array}{rr}{-1-1} & {-4} \\\ {1} & {-1-1}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
$$
\left(\begin{array}{rr}{-2} & {-4} \\\ {1} & {-2}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
Solving this system of equations, we get the eigenvector:
$$
\vec{v}_{1} = \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
For $\lambda_2 = -3,$ we have:
$$
\left(\begin{array}{rr}{-1-(-3)} & {-4} \\\ {1} & {-1-(-3)}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
$$
\left(\begin{array}{rr}{2} & {-4} \\\ {1} & {2}\end{array}\right)\left(\begin{array}{r}{x_{1}} \\ {x_{2}}\end{array}\right) = \left(\begin{array}{r}{0} \\ {0}\end{array}\right)
$$
Solving this system of equations, we get the eigenvector:
$$
\vec{v}_{2} = \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
#tag_title#Step 3: Construct the general solution#tag_content#Now that we have the eigenvalues and eigenvectors, we can construct the general solution of the given system of linear differential equations using the following formula:
$$
\vec{x}(t) = c_{1} e^{\lambda_1 t} \vec{v}_{1} + c_{2} e^{\lambda_2 t} \vec{v}_{2}
$$
Substituting the eigenvalues and eigenvectors:
$$
\vec{x}(t) = c_{1} e^{\left(1\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right) + c_{2} e^{\left(-3\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
#tag_title#Step 4: Analyze the behavior as \(t \rightarrow \infty\) #tag_content#To analyze the behavior of the general solution as \(t \rightarrow \infty\), we consider the exponential terms:
- For the first term, $e^{(1)t}$, as \(t \rightarrow \infty\), the term will grow exponentially, making the solution unbounded.
- For the second term, $e^{(-3)t}$, as \(t \rightarrow \infty\), the term will tend to zero, making its contribution in the general solution nearly negligible.
Thus, as \(t \rightarrow \infty\), the behavior of the solution is noted to grow exponentially, primarily driven by the first term. The general solution becomes unbounded.
In conclusion, the general solution of the given system of linear differential equations is:
$$
\vec{x}(t) = c_{1} e^{\left(1\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right) + c_{2} e^{\left(-3\right) t} \left(\begin{array}{r}{2} \\ {1}\end{array}\right)
$$
And as \(t \rightarrow \infty\), the solution is unbounded, growing exponentially due to the first term.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Eigenvalues
Eigenvalues play a crucial role in solving differential equations, especially when dealing with systems of linear equations. An eigenvalue is a scalar that provides insight into the behavior of a dynamical system. To find the eigenvalues of a matrix, you solve the characteristic equation, which is derived by subtracting \(\lambda\) times the identity matrix from your matrix and setting the determinant to zero. This gives us roots, \(\lambda_1\) and \(\lambda_2\), which indicate growth or decay patterns in solutions.
- Positive eigenvalues suggest exponential growth.
- Negative eigenvalues suggest exponential decay.
- Complex eigenvalues indicate oscillatory behavior.
Exploring Matrix Theory
Matrix theory is fundamental when dealing with systems of linear equations. A matrix is a rectangular array of numbers or functions, and its properties help determine the behavior of dynamic systems.
- The determinant of a matrix gives information about the invertibility of the matrix and helps find eigenvalues.
- Matrices can represent linear transformations, rotating, scaling, or translating data in various ways.
- Matrix eigenvectors, associated with eigenvalues, represent direction vectors that scale during transformation.
Visualizing Direction Fields
A direction field is a visual representation that helps us understand the behavior of differential equations without solving them analytically. For a given differential equation, it consists of small line segments drawn at various points, each indicating the slope or direction of the solution at that point.
- They help us predict long-term behavior by observing patterns and trends.
- They visually show stability or instability of equilibrium points.
- They provide a way to sketch solution trajectories based on initial conditions.
Investigating Trajectories
Trajectories are paths that solutions of a system of differential equations follow over time. In dynamical systems, trajectories trace the states a system can pass through, given an initial point.
- Trajectories can spiral outwards, converge to a point, or oscillate, depending on the eigenvalues and system's nature.
- Stable nodes represent convergent trajectories, leading to equilibrium.
- Unstable nodes indicate divergence, where trajectories move away from equilibrium.
