Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -3 & -1 & 0 \\ 1 & -1 & 0 \\ -1 & -1 & -2 \end{array}\right] \mathbf{y} $$

Short answer

Question: Determine the general solution to the system of ODEs given by the matrix: $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -3 & -1 & 0 \\ 1 & -1 & 0 \\ -1 & -1 & -2 \end{array}\right] \mathbf{y} $$ Answer: To determine the general solution to the given system, follow these steps: 1. Identify and rewrite the system of ODEs. 2. Find the eigenvalues of the matrix. 3. Find the corresponding eigenvectors for each eigenvalue. 4. Construct the general solution using the formula: \(\mathbf{y}(t)=c_1e^{\lambda_1t}\mathbf{x_1}+c_2e^{\lambda_2t}\mathbf{x_2}+c_3e^{\lambda_3t}\mathbf{x_3}\), where \(c_1, c_2, c_3\) are constants.

Step-by-step solution

Identify and rewrite the system of ODEs

The given system is: $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -3 & -1 & 0 \\ 1 & -1 & 0 \\ -1 & -1 & -2 \end{array}\right] \mathbf{y} $$ Here, \(\mathbf{y}\) is a column vector with elements: \(\mathbf{y}=\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}\). So the system consists of three ODEs: $$ \begin{cases} y_1' = -3y_1 - y_2 \\ y_2' = y_1 - y_2 \\ y_3' = -y_1 - y_2 - 2y_3 \\ \end{cases} $$

Find the eigenvalues of the matrix

To find the eigenvalues, we first find the characteristic equation of the given matrix \(A\): $$ \det (A - \lambda I) = 0 $$ Where the matrix \(A\) is: $$ \left[\begin{array}{rrr} -3 & -1 & 0 \\ 1 & -1 & 0 \\ -1 & -1 & -2 \end{array}\right] $$ And the identity matrix \(I\) is: $$ \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $$ So the characteristic equation is: $$ \det \left(\begin{array}{ccc} -3-\lambda & -1 & 0 \\ 1 & -1-\lambda & 0 \\ -1 & -1 & -2-\lambda \end{array}\right) = 0 $$ Solve it to find the eigenvalues: \(\lambda_1, \lambda_2, \lambda_3\).

Find the eigenvectors of the matrix

For each eigenvalue \(\lambda_i\), we need to find a non-trivial solution (eigenvector) to the following system: $$ (A - \lambda_i I) \mathbf{x_i} = 0 $$ Solve the systems to find the eigenvectors: \(\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}\).

Construct the general solution

Now, we can write the general solution to the system of ODEs using the eigenvalues and eigenvectors. The general solution is given by the formula: $$ \mathbf{y}(t)=c_1e^{\lambda_1t}\mathbf{x_1}+c_2e^{\lambda_2t}\mathbf{x_2}+c_3e^{\lambda_3t}\mathbf{x_3} $$ Where \(c_1, c_2, c_3\) are constants. This is the general solution to the given system of ODEs, which can be written as a column vector: $$ \mathbf{y}(t)=\begin{bmatrix} y_1(t) \\ y_2(t) \\ y_3(t) \end{bmatrix} $$

Key concepts

Eigenvalues and Eigenvectors

When dealing with a system of ordinary differential equations (ODEs), eigenvalues and eigenvectors provide a powerful method for finding the general solution. Understanding these concepts is essential for solving matrix differential equations.

Eigenvalues are particular scalars associated with a linear system of equations, and they are critical in determining the behavior of the system over time. They are derived from the characteristic equation of a matrix, which is just a determinant that sets the matrix minus a scalar multiple of the identity matrix to zero. The roots of the characteristic equation, which are the values of the scalar that satisfy the equation, are the eigenvalues.

Eigenvectors, on the other hand, are the vectors that, when multiplied by the matrix, only stretch or contract, without any rotation. In the context of solving a system of ODEs, eigenvectors correspond to the directions in which the system evolves over time. For each eigenvalue, there is at least one non-zero eigenvector that satisfies the equation \(A - \lambda I) \mathbf{x} = 0\), where \(A\) is the matrix of coefficients, \(\lambda\) is an eigenvalue, \(I\) is the identity matrix, and \(\mathbf{x}\) is the eigenvector.

In the provided exercise, solving for the eigenvalues and corresponding eigenvectors serves as the basis for constructing the solution to the differential equation. This includes solving the characteristic equation, then using each eigenvalue to find its associated eigenvector.

Characteristic Equation

The characteristic equation is the key to unlocking the behavior of a matrix-operated system. In simple terms, it's a polynomial equation that we derive from a square matrix to find its eigenvalues, the roots of this polynomial.

To form the characteristic equation of a matrix \(A\), we subtract \(\lambda\) times the identity matrix (\(I\)) from \(A\), and then take the determinant of this new matrix. The result is a polynomial in terms of \(\lambda\) that is equal to zero, which is our characteristic equation:\[\det (A - \(\lambda\) I) = 0\].

Solving the characteristic equation may require skills in algebra such as factorization or using the quadratic formula for smaller matrices, or more advanced methods for larger matrices. For the overall system's general solution, each root \(\lambda\) found from the characteristic equation becomes part of the exponential function in the final solution, providing information on how solutions to the ODE system behave as time progresses. In the example from the exercise, you would calculate the determinant of the matrix after substituting \(\lambda\) and then solve for \(\lambda\) to find the eigenvalues.

Matrix Differential Equation

Matrix differential equations involve matrices and differential equations, a natural combination in systems of linear ODEs. They take the form:\[\mathbf{y}^{\prime} = A\mathbf{y}\], where \(\mathbf{y}^{\prime}\) is the derivative of a vector function \(\mathbf{y}\), and \(A\) is a matrix. In this compact form, a single matrix equation represents multiple interconnected differential equations.

The general solution to this type of problem often involves an exponential function of a matrix. However, when the matrix has distinct eigenvalues, as in the case of our example, we can expedite the process. We find the eigenvalues and corresponding eigenvectors of \(A\) and use them to write the general solution in a form that is a linear combination of exponential terms, each multiplied by an eigenvector and a constant. This leads to \(\mathbf{y}(t) = c_1e^{\lambda_1t}\mathbf{x_1} + c_2e^{\lambda_2t}\mathbf{x_2} + c_3e^{\lambda_3t}\mathbf{x_3}\).

Each term in this solution represents a particular solution to the system evolving over time along a direction defined by its eigenvector, growing or decaying exponentially with a rate determined by the eigenvalue. For those studying the subject, mastering the process of solving matrix differential equations is an invaluable tool, enabling them to tackle a wide variety of problems in applied mathematics, physics, and engineering.