Question

Solve the given initial value problem. Describe the behavior of the solution as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-2} & {1} \\ {-5} & {4}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{1} \\ {3}\end{array}\right) $$

Short answer

In summary, we found the particular solution of the given initial value problem to be: $$ \mathbf{x}(t) = e^{t} \begin{pmatrix} 1 \\ 3 \end{pmatrix}. $$ As \(t\) approaches infinity, the solution will grow infinitely along the direction of the eigenvector, which is \(v = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\).

Step-by-step solution

Find the eigenvalues and eigenvectors of the given matrix

Given matrix: $$ A=\left(\begin{array}{cc}{-2} & {1} \\\ {-5} & {4}\end{array}\right) $$ First, we need to find the eigenvalues by solving the characteristic equation, which is the determinant of (\(A - \lambda I\)): $$ \det(A-\lambda I) = \begin{vmatrix} -2-\lambda & 1 \\ -5 & 4-\lambda \end{vmatrix} = \lambda^2 - 2\lambda + 1 = (\lambda-1)^2 $$ So there is a single eigenvalue: $$\lambda = 1.$$ Now we find the eigenvector corresponding to this eigenvalue by solving \((A - \lambda I)v = 0\): $$ \left(\begin{array}{cc}{-3} & {1} \\\ {-5} & {3}\end{array}\right) \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 $$ We can notice that the second row is just a multiple of the first row, so we just need to solve the first equation: $$ -3v_1 + v_2 = 0 \implies v_2 = 3v_1 $$ We can choose \(v_1 = 1\), so \(v_2 = 3\), and the eigenvector is: $$ v = \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$

Write the general solution of the system

The general solution of a linear ODE system with constant coefficients can be written as: $$ \mathbf{x}(t) = c e^{\lambda t} v $$ For our problem: $$ \mathbf{x}(t) = c e^{t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$

Apply the initial condition to find the particular solution

We are given the initial condition: $$ \mathbf{x}(0)=\left(\begin{array}{l}{1} \\\ {3}\end{array}\right) $$ Plugging in \(t = 0\) to our general solution, we have: $$ \begin{pmatrix} 1 \\ 3 \end{pmatrix} = c e^{0} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = c \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$ This already satisfies the given initial condition, so \(c=1\). Thus, the particular solution is: $$ \mathbf{x}(t) = e^{t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$

Analyze the behavior of the solution as \(t \rightarrow \infty\)

As \(t \rightarrow \infty\), the solution \(\mathbf{x}(t)\) will also go to infinity since the eigenvalue is positive (\(\lambda = 1> 0\)) and it's exponential with base \(e\). The solution's behavior will be dominated by the eigenvector, so it will grow along the direction of the eigenvector \(v = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\).

Key concepts

Eigenvalues and Eigenvectors

When working with linear systems of differential equations, eigenvalues and eigenvectors play a crucial role. Consider a matrix system \[ A \mathbf{x} = \lambda \mathbf{x} \]where \(A\) is a square matrix, \(\lambda\) are the eigenvalues, and \(\mathbf{x}\) are the eigenvectors. Solving for the eigenvalues involves finding the roots of the characteristic equation, which is derived from the determinant \[ \det(A - \lambda I) = 0 \]The eigenvectors are the directions along which the transformation associated with the matrix \(A\) acts as simple scaling. In simpler terms, they point towards the direction where the stretching or compressing occurs.
Eigenvalues give us essential information about system stability. In the provided solution, a single eigenvalue \(\lambda = 1\) shows that the solutions will grow exponentially as time \(t\) progresses.

Linear Systems of ODEs

Systems of linear ordinary differential equations (ODEs) are widespread in modeling real-world phenomena. Such systems can be expressed concisely in matrix form as\[ \mathbf{x}' = A \mathbf{x} \]where \(\mathbf{x}\) is a vector of functions and \(A\) is a matrix of coefficients. Solving this system involves breaking it down into simpler parts using eigenvalues and eigenvectors.
The solution is strongly dependent on the nature of the matrix \(A\) and whether it can be diagonalized. If diagonalizable, each variable component of \(\mathbf{x}\) can be solved separately. For complex or repeated eigenvalues, generalized eigenvectors come into play. Here, with a repeated eigenvalue, the solution forms a particular pattern defined mainly by the eigenvector structure.

Initial Value Problems

Initial value problems are common setups where we solve differential equations given specific conditions at \(t = 0\). For the problem \[ \mathbf{x}(0) = \begin{pmatrix} 1 \ 3 \end{pmatrix} \]we use these initial conditions to determine specific constants in the general solution. By substituting \(t = 0\) into the general solution, we identify that \[ c = 1 \]from the given system. Initial value problems help us tailor the solution specifically to ensure it fits a real-world scenario or boundary, making it practical for predictive purposes in science and engineering.

Exponential Solutions

Exponential solutions arise naturally from the differential equations due to the appearance of eigenvalues in the form of exponential functions. Solving a system typically results in solutions like \[ \mathbf{x}(t) = c e^{\lambda t} \mathbf{v} \]where \(c\) is a constant, \(\lambda\) is the eigenvalue, and \(\mathbf{v}\) is the related eigenvector. The exponential term \(e^{\lambda t}\) characterizes the growth or decay rate of the system.
If \(\lambda > 0\), solutions grow over time, as seen in this problem where \(\mathbf{x}(t)\) expands indefinitely along \(\mathbf{v}\). Conversely, if \(\lambda < 0\), solutions decay, heading towards zero. This exponential behavior is central to understanding the long-term behavior of dynamic systems modeled by ODEs.