Problem 38 - Advanced Engineering Mathematics

Question

Use diagonalization to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{ll}1 & 3 \\ 2 & 2\end{array}\right) \mathbf{X}+\left(\begin{array}{l}e^{t} \\ e^{t}\end{array}\right)\)

Step-by-Step Solution

Answer

The system can be solved using the diagonalized matrix and transforming it with the calculated inverse and eigenvectors.

Step 1: Identify the Matrix A

The given system is \( \mathbf{X}' = A\mathbf{X} + \mathbf{G}(t) \), where \( A = \begin{pmatrix} 1 & 3 \ 2 & 2 \end{pmatrix} \) is the matrix of the system coefficients.

Step 2: Find Eigenvalues of A

To diagonalize the matrix \( A \), we first find its eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. This leads to \( \det(\begin{pmatrix} 1-\lambda & 3 \ 2 & 2-\lambda \end{pmatrix}) = 0 \), which simplifies to \( (1-\lambda)(2-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = 0 \). Solving this quadratic equation gives the eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = -1 \).

Step 3: Find Eigenvectors of A

For each eigenvalue, solve \( (A - \lambda_i I)\mathbf{v}_i = \mathbf{0} \) to find the corresponding eigenvectors. For \( \lambda_1 = 4 \), solve \( (A - 4I)\mathbf{v} = \mathbf{0} \). This gives \( \begin{pmatrix} -3 & 3 \ 2 & -2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \), leading to eigenvector \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \). For \( \lambda_2 = -1 \), solve \( (A + I)\mathbf{v} = \mathbf{0} \), giving \( \begin{pmatrix} 2 & 3 \ 2 & 3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \), leading to eigenvector \( \mathbf{v}_2 = \begin{pmatrix} -3 \ 2 \end{pmatrix} \).

Step 4: Form the Matrix P and Diagonal Matrix D

Construct the matrix \( P \) using the eigenvectors obtained: \( P = \begin{pmatrix} 1 & -3 \ 1 & 2 \end{pmatrix} \). The diagonal matrix \( D \) is formed using the eigenvalues: \( D = \begin{pmatrix} 4 & 0 \ 0 & -1 \end{pmatrix} \).

Step 5: Compute \( P^{-1} \) and Transform the System

Calculate the inverse of \( P \), \( P^{-1} = \frac{1}{8}\begin{pmatrix} 2 & 3 \ -1 & 1 \end{pmatrix} \). Transform the system using \( \mathbf{Y} = P^{-1}\mathbf{X} \) to decouple the system: \( \mathbf{Y}' = D\mathbf{Y} + P^{-1}\mathbf{G}(t) \).

Step 6: Find the General Solution for \( \mathbf{Y}(t) \)

The transformed system \( \mathbf{Y}' = D\mathbf{Y} + P^{-1}\mathbf{G}(t) \) yields two decoupled equations. Solve these using standard methods for linear differential equations with non-homogeneous terms. The solutions incorporate the system eigenvalues and forcing terms in \( \mathbf{G}(t) \).

Step 7: Transform Back to Find \( \mathbf{X}(t) \)

Once the solutions for \( \mathbf{Y}(t) \) are found, transform back to the original variables using \( \mathbf{X}(t) = P\mathbf{Y}(t) \). This provides the general solution of the initial system in terms of \( \mathbf{X}(t) \).

Step 8: Simplify and Verify the Solution

Ensure that the solution satisfies the original system equations and check for consistency. Simplify the expression where possible to make it easier to interpret.

Key Concepts

Diagonalization

Diagonalization is a technique used in linear algebra that simplifies the study of linear transformations, particularly for differential equations. The process involves breaking down a complex matrix into a simpler form, where calculations become more manageable.
By diagonalizing a matrix, we turn it into a diagonal matrix composed mainly of eigenvalues on its diagonal. This makes operations such as exponentiation and solving differential equations easier since diagonal matrices are far simpler to work with.

• To start, identify a matrix \( A \) from your system of equations.
• Find the eigenvalues of \( A \) to help create our diagonal matrix.
• Construct a matrix \( P \) using the eigenvectors of \( A \). \( P \) conveys the transformation from the original to the diagonalized system.

Diagonalization is particularly useful in solving linear systems of differential equations, as it reduces the system to a set of easier, decoupled equations.

Eigenvalues and Eigenvectors

Calculating eigenvalues and eigenvectors is foundational in the diagonalization process. Eigenvalues are scalars that give insight into the matrix's properties and dynamics of the differential system.

• The characteristic equation \( \det(A - \lambda I) = 0 \) is used to find the eigenvalues of a matrix \( A \). For example, solving the equation \( \lambda^2 - 3\lambda - 4 = 0 \) yields eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = -1 \).
• Once eigenvalues are discovered, corresponding eigenvectors are found by solving \( (A - \lambda I)\mathbf{v} = \mathbf{0} \).
• For example, for eigenvalue \( \lambda_1 = 4 \), the eigenvector can be \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
• For eigenvalue \( \lambda_2 = -1 \), the eigenvector can be \( \mathbf{v}_2 = \begin{pmatrix} -3 \ 2 \end{pmatrix} \).

Eigenvalues and eigenvectors map out the direction and magnitude of the transformation, key to simplifying the original problem.

Matrix Transformation

Matrix transformation allows us to convert a system into a more solvable format. By using a transformation matrix \( P \) constructed from eigenvectors, we translate the system into one with a simple diagonal matrix \( D \).

• Construct \( P \) using the eigenvectors of \( A \): \( P = \begin{pmatrix} 1 & -3 \ 1 & 2 \end{pmatrix} \).
• The diagonal matrix \( D \) is composed of the eigenvalues: \( D = \begin{pmatrix} 4 & 0 \ 0 & -1 \end{pmatrix} \).
• Calculate \( P^{-1} \), which is necessary to transform between the original and reduced systems.

The transformation solves the system twofold: first by simplifying the system with \( \mathbf{Y}' = D\mathbf{Y} + P^{-1}\mathbf{G}(t) \), and then returning to the original setting with \( \mathbf{X}(t) = P\mathbf{Y}(t) \). This dance between bases turns intricate problems into elementary solutions.

Non-Homogeneous Differential Equations

Non-homogeneous differential equations include additional terms beside the derivative and linear component. These additional elements are called forcing terms, represented in the system as \( \mathbf{G}(t) \).

• The given system \( \mathbf{X}' = A\mathbf{X} + \mathbf{G}(t) \) showcases non-homogeneity due to \( \mathbf{G}(t) = \begin{pmatrix} e^t \ e^t \end{pmatrix} \).
• To solve, we use the diagonalized system after incorporating changes using \( P^{-1} \). This modifies the forcing terms for a simplified calculation.

In solving the transformed equations \( \mathbf{Y}' = D\mathbf{Y} + P^{-1}\mathbf{G}(t) \), the effects of the non-homogeneous \( \mathbf{G}(t) \) are straightforwardly integrated, resulting in comprehensive solutions adaptable back to original variable settings.