Question

In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. \(\mathbf{y}^{\prime}=\left[\begin{array}{lll}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}2 \\\ 5 \\ 5\end{array}\right]\) [For Exercise 8, the characteristic polynomial is \(p(\lambda)=-\lambda^{2}(\lambda-6)\).]

Short answer

Question: Find the particular solution to the given initial value problem involving a homogeneous linear system with matrix \(A = \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}\) and initial condition \(\mathbf{y}(0) = \begin{bmatrix} 2 \\ 5 \\ 5 \end{bmatrix}\). Answer: The particular solution is \(\mathbf{y}(t) = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + 3 e^{6 t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\).

Step-by-step solution

Find the eigenvalues

First, we need to find the eigenvalues of the given matrix \(A = \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}\). We are given the characteristic polynomial for this matrix: \(p(\lambda) = -\lambda^2(\lambda - 6)\). The eigenvalues are the roots of this polynomial, so we will solve the following equation: \(-\lambda^2(\lambda - 6) = 0\). We find that the eigenvalues are \(\lambda_1 = 0\), \(\lambda_2 = 0\), and \(\lambda_3 = 6\).

Find the eigenvectors

Next, we need to find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = 0\), we solve the equation \((A - \lambda_1 I)\mathbf{x} = \mathbf{0}\), that is: \(\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix} \mathbf{x} = \mathbf{0}\) The eigenvector \(\mathbf{x}_1\) is then found to be: \(\mathbf{x}_1 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}\). For \(\lambda_2 = 0\), our eigenvector is the same, so we do not need to repeat the process. For \(\lambda_3 = 6\), we solve the equation \((A - \lambda_3 I)\mathbf{x} = \mathbf{0}\), that is: \(\begin{bmatrix} -4 & 2 & 2 \\ 2 & -4 & 2 \\ 2 & 2 & -4 \end{bmatrix} \mathbf{x} = \mathbf{0}\) The eigenvector \(\mathbf{x}_3\) is then found to be: \(\mathbf{x}_3 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\).

Form the general solution

Now, we can form the general solution of the homogeneous linear system using the eigenvalues and eigenvectors. The general solution is: \(\mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{x}_1 + c_2 e^{\lambda_2 t} \mathbf{x}_2 + c_3 e^{\lambda_3 t} \mathbf{x}_3\) where \(c_1, c_2, c_3\) are scalar constants. Plugging in the eigenvalues and eigenvectors we found previously, we get: \(\mathbf{y}(t) = c_1 e^{0 t} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + c_2 e^{0 t} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + c_3 e^{6 t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\) Simplifying, we get: \(\mathbf{y}(t) = (c_1 + c_2) \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + c_3 e^{6 t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\)

Use the initial condition to find the particular solution

Finally, we can use the initial condition \(\mathbf{y}(0) = \begin{bmatrix} 2 \\ 5 \\ 5 \end{bmatrix}\) to find the particular solution. Plugging in \(t=0\) and \(\mathbf{y}(0)\) into our general solution, we get: \(\begin{bmatrix} 2 \\ 5 \\ 5 \end{bmatrix} = (c_1 + c_2) \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + c_3 e^{0} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\) Solving the system of equations, we obtain \(c_1 + c_2 = 1\) and \(c_3 = 3\). Thus, the particular solution for this initial value problem is: \(\mathbf{y}(t) = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + 3 e^{6 t} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\)

Key concepts

Eigenvalues and Eigenvectors

To fully grasp the concept of eigenvalues and eigenvectors, it's essential to understand their purpose and how they fit within the realm of linear algebra. Eigenvalues are special numbers, \(\lambda\), associated with a matrix that provide insight into the transformation represented by that matrix. These numbers are critical because they tell us how much a given vector is stretched or compressed during the transformation. Eigenvectors, on the other hand, are non-zero vectors, \(\mathbf{x}\), that only change by a scalar factor when the matrix is applied. In other words, eigenvectors maintain their direction during the transformation.The relationship between eigenvectors, eigenvalues, and a matrix \(A\) is defined by the equation \(A\mathbf{x} = \lambda\mathbf{x}\). When solved, this equation helps find the eigenvalues, \(\lambda\), which are used to determine the eigenvectors, \(\mathbf{x}\). Finding these might seem complex, but they are the foundation for solving many types of problems in physics, engineering, and computer science due to their ability to simplify matrix operations. They allow us to decompose complex systems into their basic components, making them easier to analyze.

Characteristic Polynomial

The characteristic polynomial is a fundamental tool for finding the eigenvalues of a matrix. It is a polynomial created during the process of calculating eigenvalues and serves as a bridge connecting matrices with algebraic equations.To find the characteristic polynomial of a matrix \(A\), you subtract \(\lambda\) times the identity matrix \(I\) from \(A\), and then calculate the determinant of the resulting matrix: \(\det(A - \lambda I)\). This will yield a polynomial in terms of \(\lambda\), known as the characteristic polynomial.For example, given the problem's characteristic polynomial \(p(\lambda) = -\lambda^2(\lambda - 6)\), finding the roots of this polynomial, by setting it to zero, gives us the eigenvalues. The roots of the characteristic polynomial are significant because they are the eigenvalues of the matrix. They provide critical insight into the behavior of a system governed by the matrix. Understanding how to derive and utilize the characteristic polynomial is a key skill in solving linear systems.

Initial Value Problems

In the context of differential equations, initial value problems involve finding a solution that satisfies a given set of starting conditions. These problems are quite common in modeling real-world phenomena where the initial state of a system is known, and you want to predict future behavior.To solve an initial value problem, you need not only the general solution of the differential equation but also specific information about the system at time \(t=0\). This is called the initial condition and it allows you to find the unique solution that fits this condition.For example, in the given exercise, we are given the initial condition \(\mathbf{y}(0) = \begin{bmatrix} 2 \ \ 5 \ \ 5 \end{bmatrix}\). By inserting this information into the general solution, \(\mathbf{y}(t)\), you solve for any unknown constants to pin down the particular solution to the problem. This particular solution describes the system's behavior for all future points in time. Solving initial value problems is crucial in many scientific fields, as it allows researchers to simulate and predict how systems evolve over time based on known starting conditions.