Problem 20 - Advanced Engineering Mathematics

Question

In Problems 13-32, use vaniation of parameters to solve the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} 1 & 8 \\ 1 & -1 \end{array}\right) \mathbf{X}+\left(\begin{array}{c} e^{-t} \\ t e^{t} \end{array}\right) $$

Step-by-Step Solution

Answer

The solution involves both the homogeneous and particular solutions determined by variation of parameters, combining them for the general solution.

Step 1: Find the Homogeneous Solution

First, solve the homogeneous system \( \mathbf{X}^{\prime} = \left( \begin{array}{rr} 1 & 8 \ 1 & -1 \end{array} \right) \mathbf{X} \). Find the eigenvalues by solving the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( A = \left( \begin{array}{rr} 1 & 8 \ 1 & -1 \end{array} \right) \). This gives \( \lambda^2 + \lambda - 9 = 0 \), resulting in eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \).

Step 2: Determine Eigenvectors

Find the eigenvector for each eigenvalue. For \( \lambda_1 = 3 \), solve \((A - 3I)\mathbf{v} = \mathbf{0}\), which simplifies to \(\begin{array}{c} -2 & 8 \ 1 & -4 \end{array}\mathbf{v} = \mathbf{0}\). This gives the eigenvector \(\mathbf{v_1} = \begin{bmatrix} 4 \ 1 \end{bmatrix}\). Similarly, solve for \( \lambda_2 = -3 \) to find \(\mathbf{v_2} = \begin{bmatrix} 2 \ -1 \end{bmatrix}\).

Step 3: Construct the General Solution of the Homogeneous System

Given the eigenvalues and eigenvectors, construct the general solution of the homogeneous system: \(\mathbf{X_h}(t) = c_1 e^{3t} \begin{bmatrix} 4 \ 1 \end{bmatrix} + c_2 e^{-3t} \begin{bmatrix} 2 \ -1 \end{bmatrix}\) where \(c_1\) and \(c_2\) are constants.

Step 4: Set Up Variation of Parameters

For variation of parameters, find a particular solution \( \mathbf{X_p}(t) = \begin{bmatrix} u_1(t) \ u_2(t) \end{bmatrix} \), where \( u_1(t) \) and \( u_2(t) \) satisfy the system: \( W(t) \begin{bmatrix} u_1'(t) \ u_2'(t) \end{bmatrix} = \begin{bmatrix} e^{-t} \ te^{t} \end{bmatrix} \) and \( W(t) \) is the Wronskian matrix of the fundamental solution set.

Step 5: Compute the Wronskian

Calculate the Wronskian \( W(t) \) using the fundamental solutions \( \mathbf{x_1}(t) = e^{3t}\begin{bmatrix} 4 \ 1 \end{bmatrix} \) and \( \mathbf{x_2}(t) = e^{-3t}\begin{bmatrix} 2 \ -1 \end{bmatrix} \). The Wronskian is \( \text{det} \begin{bmatrix} 4e^{3t} & 2e^{-3t} \ e^{3t} & -e^{-3t} \end{bmatrix} = -6e^0 = -6 \).

Step 6: Solve for \( u_1'(t) \) and \( u_2'(t) \)

Using \( W(t) \) and the formula \( \begin{bmatrix} u_1'(t) \ u_2'(t) \end{bmatrix} = W^{-1}(t) \begin{bmatrix} e^{-t} \ te^t \end{bmatrix} \), find \( u_1'(t) = \frac{1}{6}(e^{-t} + te^t) \) and \( u_2'(t) = \frac{1}{6}(4e^{-t} - 5te^t) \).

Step 7: Integrate to Find \( u_1(t) \) and \( u_2(t) \)

Integrate \( u_1'(t) \) and \( u_2'(t) \): \( u_1(t) = \frac{1}{6}(-e^{-t} + (t - 1)e^t) + c_3 \) and \( u_2(t) = \frac{1}{6}(-4e^{-t} - ((5t + 5)t - 6)e^t) + c_4 \), where \( c_3 \) and \( c_4 \) are integration constants that are zero for particular solutions.

Step 8: Formulate Particular Solution and General Solution

The particular solution is \( \mathbf{X_p}(t) = u_1(t) \mathbf{x_1}(t) + u_2(t) \mathbf{x_2}(t) \). Combine with the homogeneous solution to get the general solution: \( \mathbf{X}(t) = \mathbf{X_h}(t) + \mathbf{X_p}(t) \).

Step 9: Finalize the Corrected Particular Solution

Evaluate and simplify the particular solution to ensure correct form. This step involves verifying the computations and simplifying \( \mathbf{X_p}(t) \) using substitutions from previous expressions into vectors and simplifying the terms.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are crucial in solving systems of differential equations. They help us understand transformations and stability of systems. To find the eigenvalues, we use the characteristic equation obtained by calculating the determinant of \( (A - \lambda I) \), where \( A \) is our square matrix and \( \lambda \) represents an eigenvalue. For this exercise, the characteristic equation is \( \lambda^2 + \lambda - 9 = 0 \), leading to eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \). This result derives from factoring the quadratic equation.

Once we have our eigenvalues, for each, we solve the equation \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) to find the corresponding eigenvectors. For \( \lambda_1 = 3 \), \( \mathbf{v_1} = \begin{bmatrix} 4 \ 1 \end{bmatrix} \) and for \( \lambda_2 = -3 \), \( \mathbf{v_2} = \begin{bmatrix} 2 \ -1 \end{bmatrix} \). These vectors are essentially directions that remain unchanged by the linear transformation described by the matrix \( A \).

Understanding eigenvalues and eigenvectors helps us not just in composing general solutions but also in discerning properties like oscillations or exponential growth and decay in dynamic systems.

Systems of Differential Equations

A system of differential equations involves multiple equations that describe how multiple quantities change, interdependently, over time. In our exercise, the system \( \mathbf{X}^{\prime} = A\mathbf{X} + \mathbf{f}(t) \) showcases a linear system with a non-homogeneous term vector \( \mathbf{f}(t) \).

To tackle such systems, identifying a homogeneous solution is the first step. The homogeneous system \( \mathbf{X}^{\prime} = A\mathbf{X} \) ignores external forces \( \mathbf{f}(t) \) and focuses on the intrinsic system behavior. Here, once the homogeneous solution is found through eigenvalues and eigenvectors, the next step involves finding a particular solution to the entire non-homogeneous system using various methods such as undetermined coefficients or variation of parameters.

Systems of differential equations are ubiquitous in modeling real-world situations, like coupled oscillations and predator-prey models, which highlights the necessity of learning effective solution strategies.

Homogeneous Solution

A homogeneous solution is derived from the system \( \mathbf{X}^{\prime} = A\mathbf{X} \), ignoring any additional non-homogeneous terms. The solution contains terms formed from eigenvalues and eigenvectors determined previously. In our specific case, the homogeneous solution is \( \mathbf{X_h}(t) = c_1 e^{3t} \begin{bmatrix} 4 \ 1 \end{bmatrix} + c_2 e^{-3t} \begin{bmatrix} 2 \ -1 \end{bmatrix} \).

Each term of this solution represents a natural mode of the system. The constants \( c_1 \) and \( c_2 \) are determined by initial conditions provided in actual problems.

In practice, homogeneous solutions describe the inherent behavior of systems without external influences. They are critical in understanding stability and long-term trends in systems, thus serving as a cornerstone in dynamic modeling.