Question

In each exercise, the general solution of the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\) is given. Determine the coefficient matrix \(A\).\(\mathbf{y}(t)=\left[\begin{array}{l}c_{1} e^{-2 t} \\ c_{2} e^{-2 t}\end{array}\right]\)

Short answer

Based on the given general solution \(\mathbf{y}(t) = \left[\begin{array}{l} c_{1}e^{-2t}\\ c_{2}e^{-2t} \end{array} \right]\), determine the coefficient matrix \(A\) for the linear system \(\mathbf{y}^{\prime} = A\mathbf{y}\). Answer: The coefficient matrix \(A\) is not uniquely determined. However, it must satisfy the condition that the sum of the elements in each row is equal to -2 (i.e., \(a_{11} + a_{12} = -2\) and \(a_{21} + a_{22} = -2\)).

Step-by-step solution

Find the first derivatives of the given solution

To find the first derivatives, we will differentiate both components of the given solution vector \(\mathbf{y}(t)\) with respect to time \(t\). The derivatives are: $$\frac{d}{dt}(c_1 e^{-2t}) = -2c_1 e^{-2t}$$ $$\frac{d}{dt}(c_2 e^{-2t}) = -2c_2 e^{-2t}$$ So, \(\mathbf{y}^{\prime}(t) = \left[\begin{array}{l} -2c_{1}e^{-2t}\\ -2c_{2}e^{-2t} \end{array} \right]\).

Set up the system of equations using the derivative equation

We know that \(\mathbf{y}^{\prime}(t) = A\mathbf{y}(t)\). We can use the previously calculated derivatives and the given solution to create a system of equations. Our system will look like: $$ \left[ \begin{array}{l} -2c_1 e^{-2t} \\ -2c_2 e^{-2t} \end{array} \right] = A \left[ \begin{array}{l} c_1 e^{-2t} \\ c_2 e^{-2t} \end{array} \right] \\ $$

Solve the system of equations for the matrix A

Let the matrix A be denoted as \(A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\). Now, we can substitute this into our system of equations: $$ \left[ \begin{array}{l} -2c_1 e^{-2t} \\ -2c_2 e^{-2t} \end{array} \right] = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \left[ \begin{array}{l} c_1 e^{-2t} \\ c_2 e^{-2t} \end{array} \right]. $$ This will give us the following equations: $$ -2c_1 e^{-2t} = a_{11} (c_1 e^{-2t}) + a_{12} (c_2 e^{-2t}) \\ -2c_2 e^{-2t} = a_{21} (c_1 e^{-2t}) + a_{22} (c_2 e^{-2t}) $$ Simplify the equations by dividing each equation by \(e^{-2t}\). We get: $$ -2c_1 = a_{11} c_1 + a_{12} c_2 \\ -2c_2 = a_{21} c_1 + a_{22} c_2 $$ Since \(c_1\) and \(c_2\) are arbitrary constants, we can determine the values of \(a_{11}, a_{12}, a_{21},\) and \(a_{22}\) by solving the following equations: $$ a_{11} + a_{12} = -2 \\ a_{21} + a_{22} = -2 $$ As we can see here, the coefficient matrix \(A\) is not uniquely determined, as we can only determine the sum of its elements in each row. Therefore, \(A\) can take on infinitely many forms, as long as the sum of the elements in each row is equal to -2.

Key concepts

General Solution of a Linear System

When we talk about the general solution of a linear system, we refer to a formula that describes all possible solutions to a given system of linear differential equations. In the context of differential equations, the general solution includes a set of arbitrary constants, usually denoted as \(c_1, c_2, \text{etc.}\), which can be adjusted to meet specific initial conditions for the problem at hand. The general solution represents a family of functions that satisfies the differential equation.

For the given exercise, the general solution is in the form of an exponential function where \(c_1\) and \(c_2\) are the arbitrary constants. This form of solution is typical for homogeneous linear differential equations with constant coefficients. It allows the representation of the system's behavior over time and aids in understanding the dynamics of the system described by the differential equation.

It's crucial for students to recognize the importance of the general solution; it is not just a single function, but rather a whole set of functions that embody all possible behaviors of the system under study. Whether the system is modelling population growth, electrical circuits, or any other process, the general solution is the key to understanding the system's potential behaviors.

Coefficient Matrix Determination

The coefficient matrix plays a pivotal role in a system of linear differential equations. Determining the coefficient matrix, denoted as \( A \), is critical because it essentially defines the relationship between the different components of the system. In our exercise, the matrix \( A \), once found, will define how each function within the vector \( \mathbf{y}(t) \) influences the rate of change of the other components.

To determine the coefficient matrix, we take the set of first derivatives and express them in matrix form. By comparing this to the original system \( \mathbf{y}^\prime(t) = A\mathbf{y}(t) \), we can often directly infer the elements of the matrix \( A \). However, as illustrated in the exercise, sometimes we're only able to determine the sum of the row's elements rather than the individual elements themselves. This underscores a common scenario in linear algebra where multiple matrices can produce the same result when multiplied with a given vector, indicating a non-unique solution. Students should grasp that the coefficient matrix encapsulates the structure and rules that govern the system's behavior.

First Derivative Calculation

The process of calculating the first derivative is fundamental to differential calculus and, by extension, to understanding differential equations. The first derivative of a function gives us the rate at which the function's value is changing at any given point. In the context of a linear differential equation system, the first derivative of the vector function \( \mathbf{y}(t) \) provides the rate of change of each component of the system.

In our exercise, we differentiate each component of \( \mathbf{y}(t) \) with respect to time \(t\). The resulting first derivatives are crucial as they lead us to establish the relationship between the system's rate of change and the system itself, which is encoded in the coefficient matrix \(A\). Understanding and correctly performing first derivative calculations is essential for students not only to solve differential equations but also to grasp deeper concepts in physics, engineering, and other fields where change over time is a critical factor.

System of Equations

A system of equations is a set of equations with multiple variables that are all interconnected. In a linear system, these equations can be represented in matrix form, making them easier to manipulate and solve. The beauty of representing a system in matrix form is that it allows for streamlined calculations using matrix operations.

In the context of our problem, after calculating the first derivative, we set up a system of equations to relate it back to the original functions using the coefficient matrix. Solving a system of equations like this often involves finding the values of the unknowns—in our case, the elements of the matrix \(A\). Solutions to such a system can be unique, or there may be an infinite number of solutions, depending on the system's properties.

Understanding Systems with Arbitrary Constants

It's crucial for students to understand that when working with systems that include arbitrary constants (like \(c_1\) and \(c_2\) in our exercise), we are dealing with a system that describes not just one specific situation, but a whole class of potential scenarios, each corresponding to different values of these constants. It’s these arbitrary constants that give us the 'general' solution; they are the reason why the system of equations can describe many solutions at once, each applicable to different initial conditions.