Problem 18 - Advanced Engineering Mathematics

Question

Use variation of parameters to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{rr}0 & 2 \\ -1 & 3\end{array}\right) \mathbf{X}+\left(\begin{array}{c}2 \\ e^{-3 t}\end{array}\right)\)

Step-by-Step Solution

Answer

The solution combines the homogeneous and particular solutions, forming a full expression for \( \mathbf{X}(t) \).

Step 1: Find the Homogeneous Solution

First, solve the homogeneous differential equation \( \mathbf{X}^{\prime} = \begin{pmatrix} 0 & 2 \ -1 & 3 \end{pmatrix} \mathbf{X} \). Find the eigenvalues by solving the characteristic equation \( \det(\begin{pmatrix} 0 & 2 \ -1 & 3 \end{pmatrix} - \lambda \mathbf{I}) = 0 \). The eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \).

Step 2: Find Eigenvectors for the Homogeneous Solution

For \( \lambda_1 = 1 \), solve \( (\begin{pmatrix} 0 & 2 \ -1 & 3 \end{pmatrix} - \mathbf{I}) \mathbf{v_1} = \mathbf{0} \). The eigenvector corresponding to \( \lambda_1 = 1 \) is \( \mathbf{v_1} = \begin{pmatrix} 2 \ 1 \end{pmatrix} \). For \( \lambda_2 = 2 \), solve \( (\begin{pmatrix} 0 & 2 \ -1 & 3 \end{pmatrix} - 2\mathbf{I}) \mathbf{v_2} = \mathbf{0} \). The eigenvector corresponding to \( \lambda_2 = 2 \) is \( \mathbf{v_2} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Step 3: Write the General Solution for Homogeneous System

The general solution to the homogeneous system is \( \mathbf{X}_h(t) = c_1 e^{t} \begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Step 4: Use Variation of Parameters

Assume a particular solution of the form \( \mathbf{X}_p(t) = \mathbf{u}(t) \begin{pmatrix} 2e^{t} & e^{2t} \ e^{t} & e^{2t} \end{pmatrix} \). Then differentiate to obtain \( \mathbf{X'}_p(t) = \mathbf{u}'(t) \begin{pmatrix} 2e^{t} & e^{2t} \ e^{t} & e^{2t} \end{pmatrix} + \mathbf{u}(t) \mathbf{A} \begin{pmatrix} 2e^{t} \ e^{2t} \ \ e^{t} \ e^{2t} \end{pmatrix} \).

Step 5: Solve for \( \mathbf{u}(t) \)

Eliminate terms by setting \( \mathbf{u}'(t) \begin{pmatrix} 2e^{t} & e^{2t} \ e^{t} & e^{2t} \end{pmatrix} = \begin{pmatrix} 2 \ e^{-3t} \end{pmatrix} \). By differentiating, use \[ \begin{pmatrix} u_1'(t) 2e^{t} + u_2'(t) e^{2t} \ u_1'(t) e^{t} + u_2'(t) e^{2t} \end{pmatrix} = \begin{pmatrix} 2 \ e^{-3t} \end{pmatrix} \].

Step 6: Solving and Integrating \( \mathbf{u}(t) \)

After setting up and solving the equations, integrate to find \( u_1(t) \) and \( u_2(t) \). The solutions are \( u_1(t) = t - \frac{1}{15}e^{-3t} \) and \( u_2(t) = -\frac{1}{5} e^{-3t} \).

Step 7: Write Particular Solution and General Solution

Substitute \( u_1(t) \) and \( u_2(t) \) into \( \mathbf{X}_p(t) \), giving the particular solution. Then the general solution is \( \mathbf{X}(t) = \mathbf{X}_h(t) + \mathbf{X}_p(t) \).

Step 8: Final General Solution

Combine the homogeneous and particular solutions:\[ \mathbf{X}(t) = c_1 e^{t} \begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + \begin{pmatrix} (t - \frac{1}{15}e^{-3t})2e^{t} - \frac{1}{5} e^{-3t} e^{2t} \ (t - \frac{1}{15}e^{-3t}) e^{t} - \frac{1}{5} e^{-3t} e^{2t} \end{pmatrix} \].

Key Concepts

Systems of Differential Equations

In the realm of differential equations, a system refers to a set of interconnected equations that relate multiple functions to their derivatives. Such systems are pivotal in illustrating how several quantities evolve in relation to each other over time. A system of differential equations can often be represented in matrix form. For instance, in our example, the system is given by:

• \( \mathbf{X}'(t) = \mathbf{A} \mathbf{X} + \mathbf{F}(t) \)
• where \( \mathbf{X}(t) \) is a vector of functions, \( \mathbf{A} \) is a constant matrix, and \( \mathbf{F}(t) \) represents a vector of non-homogeneous terms.

Matrix notation simplifies the analysis of these equations by centralizing the transformations and operations needed. Understanding these interactions helps reveal the underlying patterns and behaviors of the systems being modeled.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are crucial in solving systems of differential equations. They provide insights into the behavior of linear equations and transformations within these systems. An eigenvalue is a scalar indicating how much the eigenvector is stretched during the transformation, while the eigenvector is a direction that remains unchanged when a linear transformation is applied.Let's see how they play a role:

• To find the eigenvalues of a matrix \( \mathbf{A} \), solve the characteristic equation: \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \).
• Once eigenvalues \( \lambda_1, \lambda_2, \ldots \) are found, substitute back to solve \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \) for eigenvectors \( \mathbf{v} \).

These pairs \( (\lambda_i, \mathbf{v}_i) \) can describe the system's modes, offering a pathway to construct solutions, especially in homogeneous systems.

Homogeneous Solutions

In differential equations, a homogeneous solution addresses the part of the system without external inputs or non-homogeneous terms. It's the solution of the system represented entirely by the matrix \( \mathbf{A} \) in \( \mathbf{X}' = \mathbf{A} \mathbf{X} \). To find the homogeneous solution:

• Find the eigenvalues and eigenvectors of \( \mathbf{A} \).
• Construct the solution combining these values in the form: \\( \mathbf{X}_h(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 + \ldots \)

For the given example, the homogeneous solution includes components formed by each eigenvalue-eigenvector pair. Such a solution encapsulates the natural dynamic behavior of the system, which can be superimposed with a particular solution to address external inputs. The full solution, which is the sum of this homogeneous part and a particular solution, can describe the complete response of the system over time.