Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -3 & -3 & 4 \\ 4 & 5 & -8 \\ 2 & 3 & -5 \end{array}\right] \mathbf{y} $$

Short answer

Question: Find the general solution of the given system of linear ordinary differential equations with the matrix $$ \left[\begin{array}{rrr} -3 & -3 & 4 \\ 4 & 5 & -8 \\ 2 & 3 & -5 \end{array}\right]. $$ Answer: The general solution of the given system is: $$ \mathbf{y}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} + c_2 e^{t} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + c_3 e^{t} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, $$ where \(c_1, c_2,\) and \(c_3\) are constants.

Step-by-step solution

1. Finding the eigenvalues

To find the eigenvalues of the matrix, we need to solve the characteristic equation, which is given by: $$ \det(A - \lambda I) = 0, $$ where \(A\) is the given matrix, \(\lambda\) are the eigenvalues, and \(I\) is the identity matrix. For our matrix, the characteristic equation is: $$ \begin{vmatrix} -3 - \lambda & -3 & 4 \\ 4 & 5 - \lambda & -8 \\ 2 & 3 & -5 - \lambda \end{vmatrix} = 0. $$ Solving this equation, we find the eigenvalues \(\lambda_1 = -1, \lambda_2 = 1,\) and \(\lambda_3 = 1\).

2. Finding the eigenvectors

Next, we find the eigenvectors corresponding to each eigenvalue by solving the following system of linear equations: $$ (A - \lambda I) \mathbf{v} = \mathbf{0}, $$ where \(\mathbf{v}\) is the eigenvector. Calculate the eigenvectors for each eigenvalue: For \(\lambda_1 = -1\): $$ \begin{pmatrix} -2 & -3 & 4 \\ 4 & 6 & -8 \\ 2 & 3 & -4 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \\ v_{13} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ The eigenvector corresponding to \(\lambda_1 = -1\) is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\). For \(\lambda_2 = 1\) and \(\lambda_3 = 1\), since these eigenvalues are equal (i.e., repeated eigenvalue), we look for two linearly independent eigenvectors. Solving the system for \(\lambda_2 = 1\): $$ \begin{pmatrix} -4 & -3 & 4 \\ 4 & 4 & -8 \\ 2 & 3 & -6 \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \\ v_{23} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ The eigenvectors corresponding to \(\lambda_2 = 1\) and \(\lambda_3 = 1\) are \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\) and \(\mathbf{v}_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).

3. Constructing the general solution

Now we can write the general solution for the system of ODEs as: $$ \mathbf{y}(t) = c_1 e^{-t} \mathbf{v}_1 + c_2 e^{t} \mathbf{v}_2 + c_3 e^{t} \mathbf{v}_3, $$ where \(c_1, c_2,\) and \(c_3\) are constants. Thus, the general solution of the given system is: $$ \mathbf{y}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} + c_2 e^{t} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + c_3 e^{t} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}. $$

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial when solving systems of differential equations. The eigenvalues are numbers that help to determine the nature of solutions to these equations. To find them, we solve the characteristic equation:\[\det(A - \lambda I) = 0.\]Here, \(A\) is the matrix from the differential equation, \(\lambda\) is a potential eigenvalue, and \(I\) is the identity matrix. Once we have the eigenvalues, which in this case are \(\lambda_1 = -1, \lambda_2 = 1\), and \(\lambda_3 = 1\), we use these to find eigenvectors. An eigenvector corresponds to each eigenvalue and satisfies the equation:\[(A - \lambda I) \mathbf{v} = \mathbf{0}.\]Eigenvectors provide direction of scaling transformation when applying the matrix \(A\) to a vector. They indicate the directions in which the transformation scales the vector without rotating it.

Systems of Linear Equations

Solving systems of linear equations is a fundamental part of finding eigenvectors. When you plug each eigenvalue back into the adjusted matrix \((A - \lambda I)\), you should solve the resulting system for its eigenvectors. This involves finding the solution to a series of equations in the form:\[\begin{pmatrix}a_{11} & a_{12} & a_{13} \a_{21} & a_{22} & a_{23} \a_{31} & a_{32} & a_{33}\end{pmatrix}\begin{pmatrix}v_{1} \v_{2} \v_{3}\end{pmatrix} = \begin{pmatrix}0 \0 \0\end{pmatrix}\]Your task is to find a non-zero vector that satisfies this equation, representing each of the eigenvectors. Here, you manually solve the matrix system, usually using techniques such as substitution or row reduction. Linear systems help us understand the eigenvectors which reflect how transformations operate across different dimensions.

Characteristic Equation

The characteristic equation is essential in finding eigenvalues. Given a matrix \(A\), the characteristic equation is formed by:\[\det(A - \lambda I) = 0.\]This equation is a polynomial where the variable \(\lambda\) represents the eigenvalues. Here, you calculate the determinant of the matrix \(A - \lambda I\), which involves finding the matrix's determinant and setting it to zero. This results in a polynomial equation in terms of \(\lambda\). Solving this polynomial gives us the eigenvalues. For the exercise, the characteristic equation provided the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 1\), \(\lambda_3 = 1\), indicating the nature of the system's solutions.

General Solution

After finding the eigenvalues and eigenvectors, we construct the general solution of the differential equation system. The general solution can be expressed as a combination of exponential functions and eigenvectors:\[\mathbf{y}(t) = c_1 e^{-\lambda_1 t} \mathbf{v}_1 + c_2 e^{-\lambda_2 t} \mathbf{v}_2 + c_3 e^{-\lambda_3 t} \mathbf{v}_3,\]where \(c_1, c_2,\) and \(c_3\) are constants that depend on initial conditions. The exponentials are derived from the eigenvalues, and each eigenvector contributes a direction along which the solution evolves over time. This method illustrates the behavior of dynamical systems and showcases how different components grow or decay over time. In this exercise, this results in:\[\mathbf{y}(t) = c_1 e^{-t} \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix} + c_2 e^{t} \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} + c_3 e^{t} \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix}.\]This represents the broad spectrum of behaviors possible within the system.