Problem 9 - Advanced Engineering Mathematics

Question

In Problems 9 and 10 , solve the given initial-value problem. \(\mathbf{X}^{\prime}=\left(\begin{array}{rr}-1 & -2 \\ 3 & 4\end{array}\right) \mathbf{X}+\left(\begin{array}{l}3 \\\ 3\end{array}\right), \quad \mathbf{X}(0)=\left(\begin{array}{r}-4 \\\ 5\end{array}\right)\)

Step-by-Step Solution

Answer

\(\mathbf{X}(t) = \begin{pmatrix}-2e^{t} - 2e^{2t} + 1 \\ 2e^t + 3e^{2t} + 1\end{pmatrix}\).

Step 1: Identify the differential equation

The given differential equation for the vector function \(\mathbf{X}\) is \[\mathbf{X}^{\prime} = \begin{pmatrix}-1 & -2 \ 3 & 4\end{pmatrix} \mathbf{X} + \begin{pmatrix}3 \ 3\end{pmatrix}.\] It involves a matrix multiplication with \(\mathbf{X}\) and an additional constant vector.

Step 2: Find the homogeneous solution

To find the homogeneous solution, solve \[\mathbf{X}^{\prime} = \begin{pmatrix}-1 & -2 \ 3 & 4\end{pmatrix} \mathbf{X}.\] Begin by finding the eigenvalues and eigenvectors of the coefficient matrix. Solving the characteristic equation \(\det(A - \lambda I) = 0\) gives the eigenvalues.

Step 3: Compute the eigenvalues

Start with the characteristic equation for \(A = \begin{pmatrix}-1 & -2 \ 3 & 4\end{pmatrix}\): \[\det\begin{pmatrix}-1-\lambda & -2 \ 3 & 4-\lambda\end{pmatrix} = (-1-\lambda)(4-\lambda) - (-2)(3).\] This simplifies to \(\lambda^2 - 3\lambda + 2 = 0\). Solving this quadratic equation, find \(\lambda_1 = 1\) and \(\lambda_2 = 2\).

Step 4: Find the eigenvectors

For \(\lambda_1 = 1\), solve \((A-I)\mathbf{v} = 0\):\[\begin{pmatrix}-2 & -2 \ 3 & 3\end{pmatrix}\begin{pmatrix}x \ y\end{pmatrix} = \begin{pmatrix}0 \ 0\end{pmatrix}.\] This results in the vector \(\mathbf{v}_1 = \begin{pmatrix}1 \ -1\end{pmatrix}\). For \(\lambda_2 = 2\), solve \((A-2I)\mathbf{v} = 0\): \[\begin{pmatrix}-3 & -2 \ 3 & 2\end{pmatrix}\begin{pmatrix}x \ y\end{pmatrix} = \begin{pmatrix}0 \ 0\end{pmatrix}.\] This results in the vector \(\mathbf{v}_2 = \begin{pmatrix}2 \ -3\end{pmatrix}\).

Step 5: Write the general solution of the homogeneous equation

Using the eigenvalues and eigenvectors, the general solution for the homogeneous differential equation is \[\mathbf{X}_h(t) = c_1\begin{pmatrix}1 \ -1\end{pmatrix}e^{t} + c_2\begin{pmatrix}2 \ -3\end{pmatrix}e^{2t},\]where \(c_1\) and \(c_2\) are arbitrary constants.

Step 6: Find the particular solution

For the non-homogeneous part, assume a particular solution of the form \(\mathbf{X}_p = \begin{pmatrix}x_p \ y_p\end{pmatrix}\). Set \(\mathbf{X}^{\prime} = 0\) to simplify to \[\begin{pmatrix}0 \ 0\end{pmatrix} = A\mathbf{X}_p + \begin{pmatrix}3 \ 3\end{pmatrix}.\] This gives rise to two equations: \(-x_p - 2y_p = -3\) and \(3x_p + 4y_p = -3\). Solve this system to find \(\mathbf{X}_p = \begin{pmatrix}1 \ 1\end{pmatrix}\).

Step 7: Form the complete solution

Combine the homogeneous and particular solutions to form the complete solution \[\mathbf{X}(t) = c_1\begin{pmatrix}1 \ -1\end{pmatrix}e^{t} + c_2\begin{pmatrix}2 \ -3\end{pmatrix}e^{2t} + \begin{pmatrix}1 \ 1\end{pmatrix}.\]

Step 8: Use initial conditions to solve for constants

Substitute \(t = 0\) and \(\mathbf{X}(0) = \begin{pmatrix}-4 \ 5\end{pmatrix}\) into the complete solution:\[-4 = c_1(1) + c_2(2) + 1\] \[5 = c_1(-1) + c_2(-3) + 1.\] Solving this system of equations gives \(c_1 = -2\) and \(c_2 = -1\).

Step 9: Write the specific solution

With the constants known, the specific solution is \[\mathbf{X}(t) = -2\begin{pmatrix}1 \ -1\end{pmatrix}e^{t} - \begin{pmatrix}2 \ -3\end{pmatrix}e^{2t} + \begin{pmatrix}1 \ 1\end{pmatrix}.\] Simplify this to obtain \[\mathbf{X}(t) = \begin{pmatrix}-2e^{t} - 2e^{2t} + 1 \ 2e^t + 3e^{2t} + 1\end{pmatrix}.\]

Key Concepts

Eigenvalues and Eigenvectors

When dealing with matrix differential equations, understanding eigenvalues and eigenvectors is crucial for solving them. An eigenvalue is a special scalar related to a matrix. It represents the factor by which the eigenvector—a non-zero vector—is scaled during transformation by the matrix. In simpler terms, when a matrix acts on an eigenvector, the vector's direction remains unchanged, although its magnitude might be altered by the eigenvalue.
To compute eigenvalues, we solve the characteristic equation \(\det(A - \lambda I) = 0\). The solutions, \(\lambda\), are our eigenvalues. For each eigenvalue, there's a corresponding eigenvector \(\mathbf{v}\) that satisfies the equation \((A - \lambda I)\mathbf{v} = 0\).

• For the matrix \(A = \begin{pmatrix} -1 & -2 \ 3 & 4 \end{pmatrix}\), we found the eigenvalues 1 and 2 by solving the characteristic equation.
• The associated eigenvectors were \(\begin{pmatrix} 1 \ -1 \end{pmatrix}\) for \(\lambda_1 = 1\) and \(\begin{pmatrix} 2 \ -3 \end{pmatrix}\) for \(\lambda_2 = 2\).

These eigenvectors form the basis for the solutions to the homogeneous part of the differential equation.

Matrix Differential Equations

Matrix differential equations involve equations where the derivatives of a vector-valued function are given in terms of matrix multiplication and vector addition. They are often written in the form \(\mathbf{X}^{\prime} = A\mathbf{X} + \mathbf{b}\), where \(A\) is a matrix, \(\mathbf{X}\) is a vector function, and \(\mathbf{b}\) is a constant vector.
In our problem, the equation is \(\mathbf{X}^{\prime} = \begin{pmatrix}-1 & -2 \ 3 & 4\end{pmatrix} \mathbf{X} + \begin{pmatrix}3 \ 3\end{pmatrix}\). To solve it, we break it into two parts:

• The homogeneous part, which ignores the constant vector \(\mathbf{b}\), leading us to focus on \(\mathbf{X}^{\prime} = A\mathbf{X}\).
• The non-homogeneous part, which considers the entire equation with \(\mathbf{b}\).

By solving for the eigenvalues and eigenvectors of \(A\), we can express the solution to the homogeneous equation as a combination of exponential solutions based on these eigenvalues.

Initial Value Problem

An initial value problem involves finding a specific solution to a differential equation that satisfies given initial conditions. It is specified by providing the value of the solution at a particular point, usually \(t = 0\). This helps us determine the arbitrary constants in the general solution.
For our specific initial value problem, we were given \(\mathbf{X}(0) = \begin{pmatrix} -4 \ 5 \end{pmatrix}\). To find the constants \(c_1\) and \(c_2\) in the homogeneous solution, we substituted \(t = 0\) into the complete solution:

• \(-4 = c_1(1) + c_2(2) + 1\)
• \(5 = c_1(-1) + c_2(-3) + 1\)

Solving these equations gave us the specific values for \(c_1\) and \(c_2\). This allowed us to write down the specific solution that perfectly fits the initial condition at \(t = 0\). This is crucial in applications because it matches the solution to real-world conditions.