Problem 17 - Advanced Engineering Mathematics

Question

In Problems, for the given linear dynamical system (taken from Exercises 10.2) (a) find the general solution and determine whether there are periodic solutions, (b) find the solution satisfying the given initial condition, and, (c) with the aid of a graphing utility, plot the solution in part (b) and indicate the direction in which the curve is traversed. $$ \begin{aligned} &x^{\prime}=x+2 y \\ &y^{\prime}=4 x+3 y, \mathbf{X}(0)=(2,-2)(\text { Problem } 1, \text { Exercises } 10.2) \end{aligned} $$

Step-by-Step Solution

Answer

The general solution is found using eigenvalues and eigenvectors; it is not periodic. The specific solution satisfying \( \mathbf{X}(0) = (2, -2) \) is obtained using initial conditions.

Step 1: Write the System of Equations in Matrix Form

Convert the given system of equations into matrix form. If \( x' = x + 2y \) and \( y' = 4x + 3y \), we can express this as \( \mathbf{X}' = A\mathbf{X} \) where the vector \( \mathbf{X} \) is \([x, y]^T\) and the matrix \(A\) is \[ \begin{bmatrix} 1 & 2 \ 4 & 3 \end{bmatrix} \].

Step 2: Find the Eigenvalues and Eigenvectors of Matrix A

To find the general solution, compute the eigenvalues \( \lambda \) of matrix \(A\) by solving the characteristic equation \( \det(A - \lambda I) = 0 \). Calculate the determinants and solve the quadratic equation to find \( \lambda_1 \) and \( \lambda_2 \). Then find the corresponding eigenvectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \).

Step 3: Construct the General Solution using Eigenvectors

The general solution is composed of the eigenvectors and exponential functions of their respective eigenvalues. If the eigenvalues and eigenvectors are \( \lambda_1, \mathbf{v}_1 \) and \( \lambda_2, \mathbf{v}_2 \), then the general solution is \( \mathbf{X}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2 \).

Step 4: Determine if There are Periodic Solutions

Check the real parts of the eigenvalues. For a solution to be periodic, all eigenvalues must be purely imaginary (zero real parts). Determine if this condition holds.

Step 5: Apply Initial Conditions to Find Specific Solution

Use the given initial condition \( \mathbf{X}(0) = (2, -2) \) to solve for constants \( c_1 \) and \( c_2 \) in the general solution. Substitute \( t = 0 \) and the initial values into the general solution equations and solve for \( c_1 \) and \( c_2 \).

Step 6: Plot the Solution and Indicate the Direction

Use a graphing utility to plot \( \mathbf{X}(t) = \begin{bmatrix} x(t) \ y(t) \end{bmatrix} \) from Step 5. The direction of traversal of the trajectory is determined by increasing \( t \); indicate this on the graph.

Key Concepts

Eigenvalues

Eigenvalues play a crucial role in understanding the behavior of linear dynamical systems. They are special numbers associated with a square matrix like our matrix \( A \) in the given problem.To find the eigenvalues, we solve the characteristic equation \[ \det(A - \lambda I) = 0 \]where \( I \) is the identity matrix and \( \lambda \) are the eigenvalues. In simpler terms, eigenvalues help us understand the scaling factor that affects the entire system.In practical contexts:

• Eigenvalues tell us about the stability of the system.
• Real parts of eigenvalues indicate whether solutions grow, decay, or neither.
• We must calculate eigenvalues to move forward in solving the dynamical system.

Matrix Form

Converting a system of equations into matrix form simplifies the analysis and solution of the system. In our exercise, the equations\[\begin{align*}x' &= x + 2y \y' &= 4x + 3y\end{align*}\]can be rewritten in matrix form as \( \mathbf{X}' = A\mathbf{X} \), where \[A = \begin{bmatrix} 1 & 2 \ 4 & 3 \end{bmatrix} \]and \( \mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix} \).This transformation is vital because:

• It organizes the system into a compact structure suitable for matrix operations, such as finding eigenvalues.
• It allows us to utilize powerful mathematical tools and techniques to analyze and solve the system efficiently.

Periodic Solutions

Periodic solutions are a specific kind of system response that repeats itself at regular intervals over time. They are crucial in many real-world applications, such as in analyzing oscillatory systems.To identify periodic solutions in our matrix problem, we analyze the eigenvalues of matrix \( A \). A solution is periodic if all eigenvalues are purely imaginary. This means the real parts of all eigenvalues must be zero.Here's how to interpret periodicity:

• If eigenvalues are imaginary and have no real part, the system undergoes continuous oscillations.
• No periodic solutions imply dynamic behavior without regular repetition.

If in our exercise, the eigenvalues possess real parts, the solutions will not be periodic.

Initial Conditions

Initial conditions specify the starting values for the variables in a dynamical system. In our problem, the given initial condition is \( \mathbf{X}(0) = (2, -2) \).Initial conditions are indispensable because:

• They define a unique solution path among the many possible solutions of a system.
• By applying these conditions, we can solve for any unknown constants, such as \( c_1 \) and \( c_2 \) in our general solution.

For example, by substituting \( t = 0 \) into our general solution equation, we derive specific values for our constants, tailoring the general solution to start exactly at \( (2, -2) \). This is crucial for accurately predicting future behavior of the system from known initial states.