Question

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

Short answer

Answer: The solution of the initial value problem is \(\mathbf{y}(t) = \frac{1}{6}e^{3t}\begin{bmatrix}6 \\ 2 \\ -2\end{bmatrix} + \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + \begin{bmatrix}0\\1\\-1\end{bmatrix}\).

Step-by-step solution

Find Eigenvalues and Eigenvectors

First, we need to find the eigenvalues and eigenvectors of the given matrix \(A = \begin{bmatrix}2 & -2 & 3\\ -4 & 4 & 3 \\ 2 & 1 & 0\end{bmatrix}\). To find the eigenvalues, we must solve the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is an identity matrix and \(\lambda\) is an eigenvalue. For the given matrix, we have: $A-\lambda I= \begin{bmatrix} 2-\lambda & -2 & 3 \\ -4 & 4-\lambda & 3 \\ 2 & 1 & -\lambda \end{bmatrix} $ After calculating the determinant we get \(\lambda^3 - 6\lambda^2 + 9\lambda=0\). Simplifying and solving for \(\lambda\), we obtain \(\lambda=0,3\). Now, let's find the eigenvectors for the eigenvalues. For \(\lambda = 3\), we have: $(A - 3I)\mathbf{v} = \begin{bmatrix} -1 & -2 & 3 \\ -4 & 1 & 3 \\ 2 & 1 & -3 \end{bmatrix} \mathbf{v} = \mathbf{0} $ After Row reducing the system, we can find the eigenvector: \(\mathbf{v_1}=c_1\begin{bmatrix}6 \\ 2 \\ -2\end{bmatrix}\), where \(c_1\) is a constant. For \(\lambda = 0\), we have: $(A - 0I)\mathbf{v} = \begin{bmatrix} 2 & -2 & 3 \\ -4 & 4 & 3 \\ 2 & 1 & 0 \end{bmatrix} \mathbf{v} = \mathbf{0} $ After row reducing the system, we can find the eigenvector: \(\mathbf{v_2} = c_2\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\) and another linearly independent eigenvector \(\mathbf{v_3}=c_3\begin{bmatrix}0\\1\\-1\end{bmatrix}\), where \(c_2\) and \(c_3\) are constants.

Finding the General Solution for the System

We can write the general solution of the system as: \(\mathbf{y}(t) = c_1 e^{3t}\mathbf{v_1} + c_2 e^{0t}\mathbf{v_2} + c_3 e^{0t}\mathbf{v_3}\)

Apply the Initial Condition

Now, let's apply the initial condition \(\mathbf{y}(0)=\begin{bmatrix}1 \\ 1 \\ 5\end{bmatrix}\) to find the values of \(c_1, c_2,\) and \(c_3\). We have: \(\begin{bmatrix}1 \\ 1 \\ 5\end{bmatrix} = c_1 e^{0}\begin{bmatrix}6 \\ 2 \\ -2\end{bmatrix} + c_2 e^{0}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + c_3 e^{0}\begin{bmatrix}0\\1\\-1\end{bmatrix}\) Simplifying the equation, we get: \(\begin{bmatrix}1 \\ 1 \\ 5\end{bmatrix} = c_1 \begin{bmatrix}6 \\ 2 \\ -2\end{bmatrix} + c_2\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + c_3\begin{bmatrix}0\\1\\-1\end{bmatrix}\) Solving the equations for \(c_1\), \(c_2\) and \(c_3\), we find \(c_1=\frac{1}{6}, c_2=1, c_3=1\).

The Solution of the System

Now we have all the constants, so we can write the solution for the initial value problem as: \(\mathbf{y}(t) = \frac{1}{6}e^{3t}\begin{bmatrix}6 \\ 2 \\ -2\end{bmatrix} + e^{0t}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + e^{0t}\begin{bmatrix}0\\1\\-1\end{bmatrix}\) Or \(\mathbf{y}(t) = \frac{1}{6}e^{3t}\begin{bmatrix}6 \\ 2 \\ -2\end{bmatrix} + \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + \begin{bmatrix}0\\1\\-1\end{bmatrix}\)

Key concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra that help us understand the properties of a matrix. In an initial value problem like we have here, eigenvalues play a crucial role in solving differential equations.
They tell us about the scaling factor that the matrix applies to its eigenvectors. To find the eigenvalues of a matrix, we solve the characteristic equation:

• This involves taking the determinant of the matrix minus lambda (\( \lambda \)) times the identity matrix and setting this equal to zero.
• For example, for the matrix \( \begin{bmatrix}2 & -2 & 3 \ -4 & 4 & 3 \ 2 & 1 & 0 \end{bmatrix} \), we find the characteristic equation: \( \lambda^3 - 6\lambda^2 + 9\lambda = 0 \).
• By solving this polynomial, we determine the eigenvalues, which are \( \lambda = 0 \) and \( \lambda = 3 \) for our problem.

Understanding these eigenvalues helps us determine how the system evolves over time, especially when combining them with eigenvectors.

Eigenvectors

Eigenvectors are vectors that remain in their span when multiplied by their corresponding eigenvalues. Each eigenvalue has one or more associated eigenvectors.
These vectors indicate the directions in which the system can evolve without changing direction, making them central to finding solutions in linear systems.

• When compute for an eigenvalue, subtract \( \lambda I \) from the matrix and solve the system \((A - \lambda I)\mathbf{v} = \mathbf{0}\) to find the associated eigenvector \(\mathbf{v}\).
• For example, for our eigenvalue \( \lambda = 3 \), we solve \((A - 3I)\mathbf{v} = \mathbf{0}\), resulting in the eigenvector \(\mathbf{v_1} = \begin{bmatrix}6 \ 2 \ -2 \end{bmatrix}\).
• Similarly, for \( \lambda = 0 \), the eigenvectors found are \( \mathbf{v_2} = \begin{bmatrix}1 \ 1 \ 0 \end{bmatrix}\) and another linearly independent eigenvector \( \mathbf{v_3} = \begin{bmatrix}0 \ 1 \ -1 \end{bmatrix}\).

Eigenvectors are essential as they build the basis for the solution of the system of differential equations.

General Solution

The general solution uses the eigenvalues and eigenvectors to construct a solution for a system of linear differential equations. It is expressed as a linear combination of exponentials multiplied by eigenvectors.
The form typically looks like:

• \( \mathbf{y}(t) = c_1e^{\lambda_1t}\mathbf{v_1} + c_2e^{\lambda_2t}\mathbf{v_2} + \ldots \)
• \(c_1, c_2, \ldots\) are constants determined based on initial conditions.

For our exercise, the general solution is written as:

• \( \mathbf{y}(t) = c_1e^{3t}\mathbf{v_1} + c_2\mathbf{v_2} + c_3\mathbf{v_3} \)

Applying the initial conditions (e.g., \( \mathbf{y}(0) = \begin{bmatrix}1 \ 1 \ 5 \end{bmatrix} \)), we find the specific values for the constants \(c_1, c_2,\) and \(c_3\). The added complexity here is to correctly choose these constants so that the solution meets the conditions given in the problem statement.

Linear Algebra

Linear algebra is the mathematical framework that supports this problem-solving method. It covers concepts like matrices, vectors, determinants, and transformations that are all employed when solving differential equations.
In this exercise, linear algebra helps us:

• Formulate the matrices and equations to compute eigenvalues and eigenvectors.
• Use row reduction techniques for matrix simplification in both finding eigenvectors and solving for the constants in the general solution.

Linear algebra provides a powerful toolkit for transforming a complex system into forms that allow us to extract and analyze the behavior of dynamical systems, as showcased in this exercise. The matrix operations involved in linear algebra form the backbone of these solutions, demonstrating its pivotal role in mathematical modeling and solution finding.