Problem 5 - Advanced Engineering Mathematics

Question

Use the method of undetermined coefficients to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{ll}4 & \frac{1}{3} \\ 9 & 6\end{array}\right) \mathbf{X}+\left(\begin{array}{r}-3 \\\ 10\end{array}\right) e^{t}\)

Step-by-Step Solution

Answer

The solution is \( \mathbf{X}(t) = c_1 e^{7t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 2 \end{pmatrix} e^t \).

Step 1: Identify the Homogeneous Solution

First, we solve the homogeneous system \( \mathbf{X}^{\prime} = A\mathbf{X} \) where \( A = \begin{pmatrix} 4 & \frac{1}{3} \ 9 & 6 \end{pmatrix} \). We find the eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \). This becomes \( \det \left( \begin{pmatrix} 4-\lambda & \frac{1}{3} \ 9 & 6-\lambda \end{pmatrix} \right) = 0 \). Expanding the determinant, we get the characteristic equation \( (4-\lambda)(6-\lambda) - 3 = 0 \) which simplifies to \( \lambda^2 - 10\lambda + 21 = 0 \).

Step 2: Compute the Eigenvalues

To find the eigenvalues, solve the quadratic equation \( \lambda^2 - 10\lambda + 21 = 0 \). Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, \ b = -10, \ c = 21 \), we find the eigenvalues \( \lambda_1 = 7 \) and \( \lambda_2 = 3 \).

Step 3: Find the Eigenvectors

For each eigenvalue, find the corresponding eigenvector by solving \( (A - \lambda I)\mathbf{v} = 0 \). For \( \lambda_1 = 7 \), \( A - 7I = \begin{pmatrix} -3 & \frac{1}{3} \ 9 & -1 \end{pmatrix} \). Solving for \( \mathbf{v} \), we find \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 3 \end{pmatrix} \). For \( \lambda_2 = 3 \), \( A - 3I = \begin{pmatrix} 1 & \frac{1}{3} \ 9 & 3 \end{pmatrix} \). Solving for \( \mathbf{v} \), we find \( \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \).

Step 4: Form the Homogeneous Solution

The homogeneous solution is \( \mathbf{X}_h(t) = c_1 e^{7t} \begin{pmatrix} 1 \ 3 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} -1 \ 1 \end{pmatrix} \), where \( c_1 \) and \( c_2 \) are arbitrary constants.

Step 5: Find the Particular Solution

Assume a particular solution of the form \( \mathbf{X}_p(t) = \mathbf{a} e^t \) where \( \mathbf{a} = \begin{pmatrix} a_1 \ a_2 \end{pmatrix} \). Substitute \( \mathbf{X}_p \) into the non-homogeneous equation, yielding \( A \mathbf{a} e^t + \mathbf{c} e^t = \mathbf{a} e^t \). This simplifies to \( (A- I)\mathbf{a} = \mathbf{c} \) where \( \mathbf{c} = \begin{pmatrix} -3 \ 10 \end{pmatrix} \). Solve \( \begin{pmatrix} 3 & \frac{1}{3} \ 9 & 5 \end{pmatrix} \begin{pmatrix} a_1 \ a_2 \end{pmatrix} = \begin{pmatrix} -3 \ 10 \end{pmatrix} \) to find \( \mathbf{a} = \begin{pmatrix} 0 \ 2 \end{pmatrix} \).

Step 6: Write the General Solution

The general solution is \( \mathbf{X}(t) = \mathbf{X}_h(t) + \mathbf{X}_p(t) \), which becomes \[ \mathbf{X}(t) = c_1 e^{7t} \begin{pmatrix} 1 \ 3 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} -1 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 2 \end{pmatrix} e^t \].

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental in understanding systems of linear equations, especially in differential equations. To solve these systems, we need to understand how a matrix, like our matrix \( A = \begin{pmatrix} 4 & \frac{1}{3} \ 9 & 6 \end{pmatrix} \), transforms vectors. This transformation is characterized by the system's eigenvalues and eigenvectors.

The process begins by deriving the eigenvalues. These are scalar values, \( \lambda \), that satisfy the equation \( \det(A - \lambda I) = 0 \). In our case, solving this equation gave us the quadratic \( \lambda^2 - 10\lambda + 21 = 0 \), which we solved using the quadratic formula to find \( \lambda_1 = 7 \) and \( \lambda_2 = 3 \).

Once eigenvalues are found, we determine eigenvectors. These are non-zero vectors that remain parallel after the transformation. For each eigenvalue, \( \lambda_i \), solve \( (A - \lambda I)\mathbf{v} = 0 \) to find the corresponding eigenvectors. In our solution, for \( \lambda_1 = 7 \), the eigenvector is \( \begin{pmatrix} 1 \ 3 \end{pmatrix} \), and for \( \lambda_2 = 3 \), it is \( \begin{pmatrix} -1 \ 1 \end{pmatrix} \).

Gathering these, the eigenvectors explain how the system evolves over time along particular directions, shaped by the eigenvalues.

Homogeneous Solutions

The homogeneous part of a differential equation is fundamental to capturing the behavior of solutions without external influences. In systems representation, it appears as \( \mathbf{X}' = A\mathbf{X} \), where \( A \) is our previous matrix. The aim here is to construct the homogeneous solution \( \mathbf{X}_h(t) \).

For our problem, this involves applying the previously found eigenvalues and eigenvectors to form a solution. The homogeneous solution combines these into linear combinations that describe all general behaviors without external forces. Here, it is described as:

• \( \mathbf{X}_h(t) = c_1 e^{7t} \begin{pmatrix} 1 \ 3 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} -1 \ 1 \end{pmatrix} \)

The terms \( c_1 \) and \( c_2 \) are constants determined by initial conditions. These exponents \( e^{7t} \) and \( e^{3t} \) reflect the rate at which those aspects of the solution grow or decay over time due to their eigenvalues.

Non-Homogeneous Differential Equations

A non-homogeneous differential equation involves additional external forces or inhomogeneities. Unlike the homogeneous ones, these account for any effect besides the system's innate dynamics. In our case, the system is modified by an added term: \( \mathbf{c}e^t \), where \( \mathbf{c} = \begin{pmatrix} -3 \ 10 \end{pmatrix} \).

This modifies our system to \( \mathbf{X}' = A\mathbf{X} + \mathbf{c}e^t \). The non-homogeneous term \( \mathbf{c}e^t \) usually represents applied forces or influences and requires more than the homogeneous solution.

Addressing these influences means noticing how they affect the trajectory over time, distinct from natural dynamics. These equations demand a combined solution that includes both particular solutions for external effects and the homogeneous solutions representing natural evolution.

Particular Solutions

Finding the particular solution is about determining a specific path through the solution space that accounts for the external influences on the system. For differential equations such as ours, it requires guessing a form of solution that matches the non-homogeneous part.

We hypothesize a form \( \mathbf{X}_p(t) = \mathbf{a}e^t \), where \( \mathbf{a} \) is a constant vector. Substituting this into the differential equation leads to conditions. Here, it boils down to clearing the equation \((A-I)\mathbf{a} = \mathbf{c} \).

• Solving \( \begin{pmatrix} 3 & \frac{1}{3} \ 9 & 5 \end{pmatrix} \begin{pmatrix} a_1 \ a_2 \end{pmatrix} = \begin{pmatrix} -3 \ 10 \end{pmatrix} \) grants \( \mathbf{a} = \begin{pmatrix} 0 \ 2 \end{pmatrix} \).

The particular solution retrieved captures how external forces guide the system's evolution over time, completing the solution picture by reflecting these added dynamics intricately.