Problem 15 - Advanced Engineering Mathematics

Question

Use variation of parameters to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{ll}3 & -5 \\ \frac{3}{4} & -1\end{array}\right) \mathbf{X}+\left(\begin{array}{r}1 \\\ -1\end{array}\right) e^{t / 2}\)

Step-by-Step Solution

Answer

The system has a general solution comprising complex exponential terms and finds specific coefficients through variation of parameters.

Step 1: Find the Homogeneous Solution Structure

First, solve the homogeneous system \( \mathbf{X}' = A\mathbf{X} \), where \[ A = \begin{pmatrix} 3 & -5 \ \frac{3}{4} & -1 \end{pmatrix}. \]To do this, find the eigenvalues and eigenvectors of matrix \( A \). Calculate the characteristic equation: \[ \det(A - \lambda I) = \begin{vmatrix} 3 - \lambda & -5 \ \frac{3}{4} & -1 - \lambda \end{vmatrix} = 0. \]

Step 2: Solve the Characteristic Equation

The determinant is: \[ (3-\lambda)(-1-\lambda) - \left(-5 \times \frac{3}{4}\right) = 0. \]Simplify and solve for \( \lambda \): \[ \lambda^2 - 2\lambda + \frac{3}{4} = 0. \]Use the quadratic formula to find \( \lambda_1 = 1 + \frac{1}{2}i \) and \( \lambda_2 = 1 - \frac{1}{2}i \).

Step 3: Find Eigenvectors

For each eigenvalue, find the corresponding eigenvectors. Solve \[ (A - \lambda_1 I) \mathbf{v} = \mathbf{0} \]and \[ (A - \lambda_2 I) \mathbf{v} = \mathbf{0}. \]This yields the eigenvectors corresponding to \( \lambda_1 \) and \( \lambda_2 \).

Step 4: Write Homogeneous Solution

The homogeneous solution is \[ \mathbf{X}_h(t) = c_1 e^{(1 + \frac{1}{2}i)t} \mathbf{v}_1 + c_2 e^{(1 - \frac{1}{2}i)t} \mathbf{v}_2, \]where \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are the eigenvectors found in Step 3.

Step 5: Determine the Particular Solution

Assume a particular solution of the form \[ \mathbf{X}_p(t) = u_1(t) \mathbf{v}_1 e^{(1 + \frac{1}{2}i)t} + u_2(t) \mathbf{v}_2 e^{(1 - \frac{1}{2}i)t}, \]where \( u_1'(t) \) and \( u_2'(t) \) must satisfy a modified system obtained by substituting \( \mathbf{X}_p(t) \) and comparing it to \( \mathbf{X}' = A\mathbf{X} + \begin{pmatrix}1 \ -1\end{pmatrix} e^{t/2} \).

Step 6: Derive Equations for Coefficients

Derive the equations by equating coefficients after substituting \( \mathbf{X}_p(t) \) into the system. Use:\[ B \begin{pmatrix} u_1' \ u_2' \end{pmatrix} = \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{t/2}, \]where \( B \) is a coefficient matrix formed from eigenvectors.

Step 7: Solve for \( u_1(t) \) and \( u_2(t) \)

Integrate the equations obtained in Step 6 to find \( u_1(t) \) and \( u_2(t) \). These functions are necessary to construct the particular solution \( \mathbf{X}_p(t) \).

Step 8: Construct the General Solution

Combine the homogeneous solution with the particular solution:\[ \mathbf{X}(t) = \mathbf{X}_h(t) + \mathbf{X}_p(t). \]Substitute the expressions for \( \mathbf{X}_h(t) \) and \( \mathbf{X}_p(t) \) from earlier steps.

Key Concepts

Eigenvalues

To understand how to determine eigenvalues, we start with a square matrix. In this exercise, the matrix \( A \) is given by: \[ A = \begin{pmatrix} 3 & -5 \ \frac{3}{4} & -1 \end{pmatrix}. \] Eigenvalues, represented by \( \lambda \), are special values that give insight into the properties of the matrix. They satisfy the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. By substituting \( A - \lambda I \) and calculating the determinant, we derive: \[ (3 - \lambda)(-1 - \lambda) - \left(-5 \times \frac{3}{4}\right) = 0. \] Solving this quadratic equation using the quadratic formula, we find two eigenvalues: \( \lambda_1 = 1 + \frac{1}{2}i \) and \( \lambda_2 = 1 - \frac{1}{2}i \). These eigenvalues are complex, which is typical for systems with oscillatory behavior. Identifying eigenvalues helps us understand the response of the system.

Eigenvectors

For each eigenvalue obtained, we need to find the corresponding eigenvector. Eigenvectors are non-zero vectors that, when multiplied by the matrix, result in a scaled version of themselves. This means that \( A\mathbf{v} = \lambda\mathbf{v} \). To find an eigenvector \( \mathbf{v} \) associated with each eigenvalue, solve the equation \( (A - \lambda I)\mathbf{v} = \mathbf{0} \). For instance, substitute \( \lambda_1 = 1 + \frac{1}{2}i \) into this equation and solve. Similarly, repeat the process for \( \lambda_2 = 1 - \frac{1}{2}i \). The resulting eigenvectors will have complex components. Typically, we express them in terms of real-valued basis vectors to make further calculations easier. These eigenvectors are crucial as they help construct the solutions for the differential system.

Homogeneous Solution

The homogeneous solution of a system is the solution to the corresponding homogeneous equation \( \mathbf{X}' = A\mathbf{X} \) without any external forcing term. To find this, we use the eigenvalues and eigenvectors obtained earlier. The general form of the homogeneous solution is: \[ \mathbf{X}_h(t) = c_1 e^{(1 + \frac{1}{2}i)t} \mathbf{v}_1 + c_2 e^{(1 - \frac{1}{2}i)t} \mathbf{v}_2, \] where \( c_1 \) and \( c_2 \) are arbitrary constants, and \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are the eigenvectors associated with \( \lambda_1 \) and \( \lambda_2 \), respectively.The functions \( e^{(1 + \frac{1}{2}i)t} \) and \( e^{(1 - \frac{1}{2}i)t} \) represent the exponential growth and oscillation of solutions over time. The homogeneous solution forms the foundational component of the overall solution.

Particular Solution

The particular solution addresses the non-homogeneous part of a differential system. Here, it solves the equation \( \mathbf{X}' = A\mathbf{X} + \begin{pmatrix}1 \ -1\end{pmatrix} e^{t/2} \) by accounting for the external term. We use the method of variation of parameters to determine it.Assume a specific form \( \mathbf{X}_p(t) = u_1(t) \mathbf{v}_1 e^{(1 + \frac{1}{2}i)t} + u_2(t) \mathbf{v}_2 e^{(1 - \frac{1}{2}i)t} \), where \( u_1(t) \) and \( u_2(t) \) are functions to be determined. By substituting this expression into the differential equation and equating coefficients, we derive auxiliary equations for \( u_1'(t) \) and \( u_2'(t) \).These equations are then integrated to find \( u_1(t) \) and \( u_2(t) \). The particular solution captures the system's response to external inputs, and when combined with the homogeneous solution, gives the complete picture of the system's behavior.