Question

Solve the initial value problem. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}-2 & -5 & -1 \\ -4 & -1 & 1 \\\ 4 & 5 & 3\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}8 \\ -10 \\ -4\end{array}\right]\)

Short answer

Based on the given initial value problem with a linear homogeneous system of ordinary differential equations (ODEs) and its initial condition, we calculated the matrix exponential of the given matrix and found the solution to the system of ODEs. After applying the initial condition, we found the particular function that satisfies the system and the initial value: \(\mathbf{y}(t) = \begin{bmatrix} -66t \\ 210t \\ -172t \end{bmatrix}\).

Step-by-step solution

Calculate the matrix exponential of the given matrix

We are given the matrix A: $\mathbf{A}=\left[\begin{array}{rrr}-2 & -5 & -1 \\\ -4 & -1 & 1 \\\ 4 & 5 & 3\end{array}\right]$ We need to find the matrix exponential, denoted as e^(At). First, we need to find the eigenvalues and eigenvectors of the matrix A.

Find the eigenvalues of the matrix A

To find the eigenvalues of \(\mathbf{A}\), we must solve the characteristic equation of A, which is: \( det(\mathbf{A} - \lambda \mathbf{I}) = 0 \) where \(\lambda\) represents an eigenvalue and \(\mathbf{I}\) is the identity matrix. The equation becomes: \( det(\begin{bmatrix} -2-\lambda & -5 & -1 \\ -4 & -1-\lambda & 1 \\ 4 & 5 & 3-\lambda \end{bmatrix}) = 0 \) Computing the determinant, we get the characteristic equation: \((-\lambda^3 + 4\lambda^2 - 5\lambda - 14) = 0\) Solving the equation, we get the eigenvalues: \(\lambda_1 = -1\), \(\lambda_2 = 2\), \(\lambda_3 = -7\)

Find the eigenvectors corresponding to the eigenvalues

To find the eigenvectors of A, we will solve the system \((\mathbf{A} - \lambda_i \mathbf{I})\mathbf{v}_i = \mathbf{0}\), where \(\lambda_i\) is an eigenvalue and \(\mathbf{v}_i\) is the corresponding eigenvector. Let's find the eigenvectors for each eigenvalue: For \(\lambda_1=-1\), we get: \( (\mathbf{A} - \lambda_1 \mathbf{I})\mathbf{v}_1 = (\begin{bmatrix} -1 & -5 & -1 \\ -4 & 0 & 1 \\ 4 & 5 & 4 \end{bmatrix})\mathbf{v}_1 = \mathbf{0}\) Solving this system, we find the eigenvector \(\mathbf{v}_1 =\begin{bmatrix} 5 \\ 2 \\ -4 \end{bmatrix} \). For \(\lambda_2=2\), we get: \( (\mathbf{A} - \lambda_2 \mathbf{I})\mathbf{v}_2 = (\begin{bmatrix} -4 & -5 & -1 \\ -4 & -3 & 1 \\ 4 & 5 & 1 \end{bmatrix})\mathbf{v}_2 =\mathbf{0} \) Solving this system, we find the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}\). For \(\lambda_3=-7\), we get: \( (\mathbf{A} - \lambda_3 \mathbf{I})\mathbf{v}_3 = (\begin{bmatrix} 5 & -5 & -1 \\ -4 & 6 & 1 \\ 4 & 5 & 10 \end{bmatrix})\mathbf{v}_3 =\mathbf{0}\) Solving this system, we find the eigenvector \(\mathbf{v}_3 =\begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix} \).

Calculate the matrix exponential of A

Now that we have the eigenvalues and their corresponding eigenvectors, we can find the matrix exponential e^(At), which is given by: \(e^{\mathbf{At}} = \sum_{k = 0}^{\infty} \frac{(\mathbf{A}t)^k}{k!} = \mathbf{P} e^{\mathbf{Dt}} \mathbf{P}^{-1}\), where \(\mathbf{P}\) is the matrix of eigenvectors, \(\mathbf{D}\) is the diagonal matrix of eigenvalues, and \(\mathbf{P}^{-1}\) is the inverse of \(\mathbf{P}\). Let's find the matrices \(\mathbf{P}\), \(\mathbf{D}\), and \(\mathbf{P}^{-1}\): \(\mathbf{P} = \begin{bmatrix} 5 & 1 & -2 \\ 2 & -1 & -3 \\ -4 & -1 & 1 \end{bmatrix}\), \(\mathbf{D} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -7 \end{bmatrix}\) Calculating the inverse of \(\mathbf{P}\) using any method (e.g., row reduction, adjugate): \(\mathbf{P}^{-1} = \begin{bmatrix} -7/27 & -2/27 & -5/27 \\ 7/27 & -11/27 & 2/27 \\ -1/27 & 1/27 & -8/27 \end{bmatrix}\) To find \(e^{\mathbf{Dt}}\), calculate \(e^{\lambda_{i}t}\) for the diagonal elements: \(e^{\mathbf{Dt}} = \begin{bmatrix} e^{-t} & 0 & 0 \\ 0 & e^{2t} & 0 \\ 0 & 0 & e^{-7t} \end{bmatrix}\) Finally, let's compute the matrix exponential \(e^{\mathbf{At}} = \mathbf{P} e^{\mathbf{Dt}} \mathbf{P}^{-1}\): \(e^{\mathbf{At}} = \begin{bmatrix} -2 & 3 & 5 \\ 14 & -7 & -7 \\ -5 & -8 & -13 \end{bmatrix}t\)

Calculate the solution to the system of ODEs

Now that we have \(e^{\mathbf{At}}\), we can find the solution for the system of ODEs. Given that \(\mathbf{y}^{\prime} = \mathbf{A}\mathbf{y}\) and the initial condition \(\mathbf{y}(0) = \begin{bmatrix} 8 \\ -10 \\ -4 \end{bmatrix}\), the solution to the system of ODEs is: \(\mathbf{y}(t) = e^{\mathbf{At}} \mathbf{y}(0) = \begin{bmatrix} -2 & 3 & 5 \\ 14 & -7 & -7 \\ -5 & -8 & -13 \end{bmatrix}t \begin{bmatrix} 8 \\ -10 \\ -4 \end{bmatrix}\) Calculating the matrix-vector product, we get: \(\mathbf{y}(t) = \begin{bmatrix} -16 - 30 - 20 \\ 112 + 70 + 28 \\ -40 - 80 - 52 \end{bmatrix}t\) Therefore, the solution is: \(\mathbf{y}(t) = \begin{bmatrix} -66t \\ 210t \\ -172t \end{bmatrix}\).

Key concepts

Matrix Exponential

When we talk about the matrix exponential, it is an essential concept in solving systems of linear differential equations. The matrix exponential, denoted as \( e^{\mathbf{A}t} \), transforms a square matrix \( \mathbf{A} \) to address systems over time. To compute this expression, we often use the eigenvalues and eigenvectors of the matrix.

• The eigen-decomposition approach leverages these eigenvalues and eigenvectors to simplify the calculation of \( e^{\mathbf{A}t} \).
• This involves creating a diagonal matrix \( \mathbf{D} \) composed of eigenvalues and a matrix \( \mathbf{P} \) comprising corresponding eigenvectors.
• Thus, \( e^{\mathbf{A}t} = \mathbf{P} e^{\mathbf{Dt}} \mathbf{P}^{-1} \), where \( e^{\mathbf{Dt}} \) is straightforward since \( \mathbf{D} \) is diagonal.

By understanding these steps, finding the matrix exponential becomes simpler, allowing us to effectively solve complex systems of differential equations.

Eigenvalues and Eigenvectors

To deeply understand matrix behavior, we dive into eigenvalues and eigenvectors. These components offer insight into the matrix's structure and transformation capabilities.

• Eigenvalues (\( \lambda \)): These are solutions to the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). In simpler terms, they scale eigenvectors during matrix transformation.
• Eigenvectors (\( \mathbf{v} \)): Associated with each eigenvalue, an eigenvector signifies a direction that remains invariant (up to a scaling factor) under the influence of the matrix \( \mathbf{A} \).

The process involves solving systems of linear equations \((\mathbf{A} - \lambda_i \mathbf{I})\mathbf{v}_i = \mathbf{0}\) for each eigenvalue \( \lambda_i \). Remarkably, this seemingly abstract concept forms the backbone of determining behaviors and solutions for differential systems.

System of Differential Equations

A System of Differential Equations comprises multiple equations that involve unknown functions and their derivatives. Solving these systems typically involves finding functions that satisfy all the outlined equations simultaneously.

• The task at hand often involves setting up an initial value problem, where initial conditions guide the solution trajectory.
• For linear systems expressed as \( \mathbf{y}' = \mathbf{A} \mathbf{y} \), solutions can be computed using the matrix exponential \( e^{\mathbf{A}t} \).
• These solutions consider both the inherent properties of the system (defined by \( \mathbf{A} \)) and external conditions (the initial state \( \mathbf{y}(0) \)).

Understanding the interaction between the matrix and its initial conditions enables the construction of solutions that precisely describe the system's evolution over time.