Question

In Exercises \(1-16\) find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} 5 & -6 \\ 3 & -1 \end{array}\right] \mathbf{y} $$

Short answer

The general solution of the given system of differential equations, \(\mathbf{y}^{\prime} = A\mathbf{y}\), with \(A = \begin{bmatrix} 5 & -6 \\ 3 & -1 \end{bmatrix}\) is: \(\mathbf{y}(t) = C_1 e^{t} \begin{bmatrix} 1 \\ \frac{2}{3} \end{bmatrix} + C_2 e^{3t} \begin{bmatrix} 2 \\ 1 \end{bmatrix}\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Find the eigenvalues of matrix A

To find the eigenvalues, we will solve the characteristic equation of matrix A which is given by \(\text{det}(A - \lambda I) = 0\). In this case, the characteristic equation is: $$ \text{det} \begin{bmatrix} 5-\lambda & -6 \\ 3 & -1-\lambda \end{bmatrix} = (5-\lambda)(-1-\lambda) - (-6)(3) = \lambda^2 - 4\lambda + 3 = 0. $$ Now, we will find the roots of this quadratic equation using the quadratic formula: $$ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $$ where \(a=1, b=-4\) and \(c=3\). Evaluating, we get: $$ \lambda = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)} = \frac{4 \pm \sqrt{4}}{2} = 2 \pm 1. $$ Thus, we have two eigenvalues, \(\lambda_1 = 1\) and \(\lambda_2 = 3\).

Find the eigenvectors corresponding to each eigenvalue

For each eigenvalue, we will find the eigenvectors by solving the linear system \((A - \lambda I)\mathbf{v} = \mathbf{0}\). Eigenvalue \(\lambda_1 = 1\): $$ \begin{bmatrix} 5-1 & -6 \\ 3 & -1-1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 4 & -6 \\ 3 & -2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Subtracting two times the first row from the second row to simplify the system, we get: $$ \begin{bmatrix} 4 & -6 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. $$ Thus, we have \(4x_1 - 6x_2 = 0\), and the eigenvector corresponding to \(\lambda_1 = 1\) is \(\mathbf{v}_1 = \begin{bmatrix}1\\ \frac{2}{3}\end{bmatrix}\). Eigenvalue \(\lambda_2 = 3\): $$ \begin{bmatrix} 5-3 & -6 \\ 3 & -1-3 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 2 & -6 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Subtracting the second row from the first row to simplify the system and dividing the first row by two: $$ \begin{bmatrix} 1 & -2 \\ 0 & -2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. $$ Thus, we have \(x_1 - 2x_2 = 0\), and the eigenvector corresponding to \(\lambda_2 = 3\) is \(\mathbf{v}_2 = \begin{bmatrix}2\\ 1\end{bmatrix}\).

Construct the general solution using eigenvalues and eigenvectors

Now that we have found the eigenvalues and the corresponding eigenvectors, we can construct the general solution of the system of differential equations: $$ \mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{v}_1 + C_2 e^{\lambda_2 t} \mathbf{v}_2 = C_1 e^{t} \begin{bmatrix} 1 \\ \frac{2}{3} \end{bmatrix} + C_2 e^{3t} \begin{bmatrix} 2 \\ 1 \end{bmatrix}, $$ where \(C_1\) and \(C_2\) are constants.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is essential in various fields of mathematics and engineering, particularly when dealing with differential equations. These concepts are intertwined; eigenvalues are special numbers associated with a square matrix, while eigenvectors are non-zero vectors that change by only a scalar factor when that linear transformation is applied.

When solving the system of differential equations, we discern the matrix representing the system and check how it transforms certain vectors, the eigenvectors. This transformation strictly involves scaling these vectors by their corresponding eigenvalues. In the example given, eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \) indicate by what multiples the eigenvectors are scaled during the transformation. This property is particularly useful in systems of differential equations as it simplifies how we view the dynamics of the system over time.

The eigenvectors corresponding to these eigenvalues, \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) in the context of the problem, form the basis vectors in the solution space. This means any vector in our system can be expressed as a combination of these eigenvectors, enabling us to construct general solutions to complex problems.

Characteristic Equation

The characteristic equation is a fundamental component in finding a matrix’s eigenvalues, and thus it’s instrumental in solving differential equations. To construct this equation, one subtracts \( \lambda \) times the identity matrix from the matrix in question and sets the determinant of the resulting matrix to zero. In the provided exercise, the characteristic equation \( \lambda^2 - 4\lambda + 3 = 0 \) was obtained from such a process.

The roots of the characteristic equation are the eigenvalues of the matrix. By applying the quadratic formula or factorization, students can find the specific eigenvalues that will determine the behavior of the system's solutions. This step is crucial because it lays the groundwork for understanding the system's dynamics by revealing how the system evolves over time based on the eigenvalues acquired from the characteristic equation.

System of Differential Equations

A system of differential equations consists of multiple equations that relate functions and their derivatives. In our example, the system is represented in matrix form. This compact notation allows for a more straightforward approach to finding the general solution to the system, particularly when it can be represented as a linear combination of exponential functions scaled by constants and multipliers of the corresponding eigenvectors, as demonstrated in the final step of the solution.

The solution to the system \( \mathbf{y}(t) \) is a vector function of time, composed of linear combinations of terms involving eigenvalues, eigenvectors, and exponentials of the product of eigenvalues and time. These combinations are weighted by arbitrary constants which can later be determined by initial conditions. Such a structure becomes particularly elegant and manageable when the matrix involved has distinct eigenvalues, allowing for the system to be succinctly described using just exponential growth or decay processes aligned with the directions dictated by the eigenvectors.

In summary, the general solution showcases how a system evolves from any initial state, with the eigenvalues determining the rate of change and the direction influenced by the eigenvectors. This provides a powerful method for predicting and understanding complex dynamic systems.