Question

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

Short answer

Question: Find the general solution of the given system of linear differential equations. System of linear differential equations: $$ \begin{cases} \frac{dy_1}{dt} = -y_2\\ \frac{dy_2}{dt} = y_1 -2y_2 \end{cases} $$ Answer: The general solution to the given system of linear differential equations is: $$ \mathbf{y}(t) = c_1 e^{-t}\begin{bmatrix}1\\1\end{bmatrix} $$ where $c_1$ is a constant.

Step-by-step solution

Find Eigenvalues

To find the eigenvalues, we need to solve the characteristic equation which is given by the determinant of (A - λI), where A is the coefficient matrix, λ are the eigenvalues and I is the identity matrix: $$ \det \left( \begin{bmatrix} 0-\lambda & -1 \\ 1 & -2-\lambda \end{bmatrix} \right) = 0 $$ Expanding the determinant: $$ (-\lambda)(-2-\lambda) - (-1)(1)=\lambda^2 + 2\lambda + 1= (\lambda +1)^2 = 0 $$ Hence, the eigenvalue is: $$ \lambda = -1 $$

Find Eigenvectors

To find the eigenvectors, we need to solve the equation (A - λI)X = 0, where X are the eigenvectors. In this case, for λ = -1, we have: $$ \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ We can solve this system of linear equations by writing the augmented matrix and row reducing it to the row-reduced echelon form: $$ \left[\begin{array}{cc|c} 1 & -1 & 0 \\ 1 & -1 & 0 \end{array}\right] \xrightarrow{{R_2 \leftarrow R_2 - R_1}}{\left[\begin{array}{cc|c} 1 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right]} $$ From the row-reduced echelon form, we have one free variable x2 = t, where t is any real number. Therefore, the eigenvector corresponding to λ = -1 is: $$ \mathbf{x} = t\begin{bmatrix}1\\1 \end{bmatrix} $$

Construct the General Solution

Now that we have the eigenvector and its corresponding eigenvalue, we can construct the general solution of the system of linear differential equations as follows: $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{x}_1 = c_1 e^{-1t}\begin{bmatrix}1\\1\end{bmatrix} $$ where c_1 is a constant.

Key concepts

Eigenvalues

In the realm of linear algebra, an **eigenvalue** is a special number associated with a square matrix. It provides significant insight into the properties of the matrix, especially when dealing with linear transformations. To find the eigenvalues of a matrix, you solve the characteristic equation.

• To determine the eigenvalues, you start by subtracting λ times the identity matrix from the original matrix.
• You then calculate the determinant of this new matrix.
• Finally, solve the resulting polynomial equation, often a quadratic, to find the values of λ.

In our problem, the matrix's characteristic polynomial was evaluated to get the eigenvalue, λ = -1. This unique eigenvalue implies special dynamics in the differential equation's solutions.

Eigenvectors

An **eigenvector** is a non-zero vector that only changes by a scalar factor when a linear transformation is applied. When paired with an eigenvalue, it gives directions along which the transformation acts by stretching or compressing, but not rotating.

• You find eigenvectors by solving the equation (A - λI)X = 0, where A is your matrix, λ is the eigenvalue, and X is the eigenvector.
• This typically involves reducing the matrix to row-echelon form and solving the resulting system of linear equations.
• For our system, λ = -1 led to an eigenvector format t\(\begin{bmatrix}1\1 \end{bmatrix} \), which indicates that every solution is a scalar multiple of the vector \(\begin{bmatrix}1\1 \end{bmatrix} \).

This illustrates that all solutions are aligned along this direction, highlighting a key aspect of the system’s behavior.

Linear Algebra

**Linear algebra** forms the backbone of understanding systems of linear equations and transforms, like our differential equation. It gives us the tools to analyze matrices, vectors, and their interactions.

• Concepts like eigenvalues and eigenvectors are integral to this discipline, simplifying the study of systems that can be described by matrices.
• Diagonalizing matrices, row reducing systems, and solving for characteristic equations are key techniques used thanks to linear algebra.
• Linear algebra not only aids in solving differential equations but also extends to numerous applications in various scientific fields, including machine learning, physics, and computer graphics.

In our exercise, linear algebra techniques guide us to the general solution of the differential system by handling the matrix operations involved in finding eigenvalues and eigenvectors.

General Solution

A **general solution** of a system of differential equations represents all possible solutions to the system. It considers each eigenvalue and corresponding eigenvector to construct a solution that describes the system's behavior.

• The general solution typically involves constants that can be adjusted to fit initial conditions.
• In the context of our given system, the solution is \(\mathbf{y}(t) = c_1 e^{-t}\begin{bmatrix}1\1\end{bmatrix}\), where \( c_1 \) is an arbitrary constant.
• This reflects a single eigensolution due to the repeated eigenvalue, expressing exponential growth or decay along the direction of its eigenvector.

Understanding the general solution gives insight into the tendencies and long-term behavior of the system, crucial for predicting future system states.