Question

Find the general solution of each of the following systems by the method of eigenvalues. (a) \(y_{1}^{\prime}=6 y_{1}-7 y_{2}\), \(y_{2}^{\prime}=y_{1}-2 y_{2}\) (b) \(\quad y_{1}^{\prime}=y_{1}+y_{2}-2 y_{3}\), \(y_{2}^{\prime}=-y_{1}+2 y_{2}+y_{3}\) \(y_{3}^{\prime}=y_{2}-y_{3}\) (c) \(y_{1}^{\prime}=2 y_{1}+y_{3}\) \(y_{2}^{\prime}=y_{2}\) $y_{3}^{\prime}=y_{1}+2 y_{3} .$$$ \begin{array}{l} \text { (d) } y_{1}^{\prime}=2 y_{1}-y_{2} \text { , }\\\ y_{2}^{\prime}=2 y_{1}+4 y_{2} .\\\ \begin{array}{ll} \text { (e) } & y_{1}^{\prime}=y_{1}-y_{2}, \\ & y_{2}^{\prime}=5 y_{1}-3 y_{2} . \\ \text { (f) } & y_{1}^{\prime}=3 y_{1}+y_{2}-y_{3}, \\ & y_{2}^{\prime}=y_{1}+3 y_{2}-y_{3}, \\ & y_{3}^{\prime}=3 y_{1}+3 y_{2}-y_{3} . \\ \text { (g) } & y_{1}^{\prime}=y_{3}, \\ & y_{2}^{\prime}=y_{1}-3 y_{3}, \\ & y_{3}^{\prime}=y_{2}+3 y_{3} . \end{array} \end{array} $$

Short answer

Answer: The general solution for the given system is: $$Y = c_1e^{t}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + c_2e^{2t}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + c_3e^{2t}\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},$$ where \(c_1\), \(c_2\), and \(c_3\) are arbitrary constants.

Step-by-step solution

1. Write the system in matrix form

The given system can be written in matrix form as: $$\begin{pmatrix} y_{1}^{\prime} \\ y_{2}^{\prime} \\ y_{3}^{\prime} \end{pmatrix} = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix}\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix}.$$ So, the matrix A corresponding to given system is: $$A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix}.$$

2. Find the eigenvalues of the matrix

To find the eigenvalues, we must solve the following equation: $$\det(A - \lambda I) = 0,$$ where \(\lambda\) is the eigenvalue and I is the identity matrix. Plugging in the given matrix A, we get: $$\det\begin{pmatrix} 2-\lambda & 0 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 0 & 2-\lambda \end{pmatrix} = 0.$$ Expanding the determinant: $$(2-\lambda)((1-\lambda)(2-\lambda)) = 0.$$ Solving for \(\lambda\), we get the eigenvalues: $$\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 2.$$

3. Determine the eigenvectors associated with each eigenvalue

Now, we will find the eigenvectors corresponding to each eigenvalue. Let's start with \(\lambda_1 = 1\): $$(A-\lambda_1 I)v_1 = 0,$$ $$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$ From the resulting equations, we get one linearly independent eigenvector: $$v_1 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}.$$ Next, let's find the eigenvectors for \(\lambda_2 = 2\): $$(A-\lambda_2 I)v_2 = 0,$$ $$\begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\\\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$ From the resulting equations, we get one linearly independent eigenvector: $$v_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}.$$ Since we have a repeated eigenvalue \(\lambda_3 = 2\), we need to find if there is a second linearly independent eigenvector. Using the same matrix as before \((A-\lambda_3 I)\), we find another linearly independent eigenvector: $$v_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.$$

4. Write and simplify the general solution using the eigenvalues and eigenvectors

Now that we have the eigenvalues and the eigenvectors, we can write the general solution as follows: $$Y = c_1e^{\lambda_1 t}v_1 + c_2e^{\lambda_2 t}v_2 + c_3e^{\lambda_3 t}v_3,$$ where \(c_1\), \(c_2\), and \(c_3\) are arbitrary constants. Plugging in the eigenvalues and eigenvectors found: $$Y = c_1e^{t}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + c_2e^{2t}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + c_3e^{2t}\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.$$ This gives us the general solution for the given system (c).

Key concepts

Method of Eigenvalues

When we deal with systems of differential equations, the method of eigenvalues is a powerful technique for finding general solutions. This method is based on the concept of eigenvalues and eigenvectors, which stem from linear algebra. The ability to express a system in matrix form lets us employ these algebraic tools to solve dynamic systems that model real-world phenomena such as oscillations and population dynamics.

The eigenvalues of a matrix are numbers that give us vital information about the behavior of the system: they can tell us whether the solutions grow, decay, or oscillate. The corresponding eigenvectors determine the direction in which these behaviors occur. To use the method of eigenvalues, we first write the system as \( AY = Y' \), where \( A \) is the matrix of coefficients, and \( Y \) is the vector of unknown functions. After calculating the eigenvalues and eigenvectors of \( A \), we can form the general solution to the system. This method simplifies what would otherwise be a complex integration problem into an algebraic one, making complex systems more approachable.

Eigenvectors and Eigenvalues

In the context of a system of differential equations, the concepts of eigenvectors and eigenvalues are critical. An eigenvalue is a scalar that, when multiplied by an eigenvector, achieves the same result as when that eigenvector is transformed by a given matrix. In simpler terms, an eigenvector of a matrix keeps its direction after being transformed by the matrix and the eigenvalue tells us how much it stretches or shrinks.

Mathematically, for a given square matrix \( A \) and non-zero vector \( v \), if \( Av = \lambda v \), then \( v \) is an eigenvector and \( \lambda \) is its corresponding eigenvalue. This definition forms the foundation for finding the general solution of linear differential systems. Because different eigenvalues correspond to independent behaviors within the system, their eigenvectors provide the essential directions for these behaviors, making them the building blocks of the general solution.

Matrix Representation of Systems

The matrix representation of systems involves expressing systems of linear differential equations as matrix equations. This form not only streamlines the calculation but also opens the door to algebraic methods for solving systems. We consolidate the coefficients of the system's equations into one square matrix and the variables into a vector.

For instance, a system like \( y_{1}' = ay_{1} + by_{2} \) and \( y_{2}' = cy_{1} + dy_{2} \) can be represented as \( Y' = AY \) where \( Y' \) is the derivative vector \( (y_{1}', y_{2}')' \) and \( A \) is the coefficient matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \). Think of the matrix as a recipe that tells us how each variable in the system interacts with the others. In terms of algebra, these interactions are linear transformations, the study of which is rich with methods for analysis and solution.

Determinant Calculation for Eigenvalues

To find the eigenvalues of a matrix associated with a system of differential equations, we perform a determinant calculation. Specifically, we start with the equation \( \det(A - \lambda I) = 0 \), where \( A \) is our coefficient matrix, \( I \) is the identity matrix, and \( \lambda \) is the scalar eigenvalue. By subtracting \( \lambda I \) from \( A \) and calculating the determinant of the resulting matrix, we get a polynomial in terms of \( \lambda \), known as the characteristic polynomial.

The roots of this polynomial are the eigenvalues of the matrix \( A \). Once we have these eigenvalues, we can plug them back into the equation \( (A - \lambda I)v = 0 \) to compute the corresponding eigenvector \( v \) for each eigenvalue. It’s like solving a puzzle: the determinant provides the clues (eigenvalues), and those clues help us find the missing pieces (eigenvectors) to complete the picture (general solution of the system).