Question

In each exercise, the general solution of the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\) is given. Determine the coefficient matrix \(A\).\(y_{1}(t)=c_{1} e^{-t}(1+2 t)+4 c_{2} t e^{-t}\) \(y_{2}(t)=-c_{1} t e^{-t}+c_{2} e^{-t}(1-2 t)\)

Short answer

Question: Determine the coefficient matrix A of the given linear system of first order differential equations, where the general solutions are given by: \(y_{1}(t)=c_{1} e^{-t}(1+2 t)+4 c_{2} t e^{-t}\) \(y_{2}(t)=-c_{1} t e^{-t}+c_{2} e^{-t}(1-2 t)\) Answer: The coefficient matrix A is: \(A = \begin{pmatrix} -1 & -3 \\ 1 & -1 \end{pmatrix}\)

Step-by-step solution

Compute Derivatives of \(y_1(t)\) and \(y_2(t)\) with Respect to \(t\)

We compute the derivatives of \(y_1(t)\) and \(y_2(t)\): \(y_1'(t)=\frac{d}{dt}(c_{1} e^{-t}(1+2 t)+4 c_{2} t e^{-t})\) \(y_2'(t)=\frac{d}{dt}(-c_{1} t e^{-t}+c_{2} e^{-t}(1-2 t))\)

Express Derivatives Using \(y_1(t)\) and \(y_2(t)\) Equations

Now we express the derivatives in terms of \(y_1(t)\) and \(y_2(t)\): \(y_1'(t)=c_{1} e^{-t}(-1+2 t)+4 c_{2} e^{-t} (1-t)\) \(y_2'(t)=-c_{1} e^{-t}(1+t)-2 c_{2} e^{-t}(1-2 t)\)

Write the Linear System \(\mathbf{y}^{\prime}=A \mathbf{y}\)

We can now write the linear system of differential equations, where \(\mathbf{y}^{\prime}\) is the vector of first derivatives and \(\mathbf{y}\) is the vector of dependent variables: \(\begin{pmatrix} y_1'(t)\\ y_2'(t) \end{pmatrix}=A \begin{pmatrix} y_1(t)\\ y_2(t) \end{pmatrix}\)

Derive Coefficients of the Matrix \(A\)

We need to find coefficients \(a_{11}\), \(a_{12}\), \(a_{21}\), and \(a_{22}\) in the matrix \(A\): \(A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\) By comparing the coefficients of \(c_1\) and \(c_2\) in the expressions for \(y_1'(t)\), \(y_2'(t)\), and \(\mathbf{y}^{\prime}=A \mathbf{y}\), we can derive the coefficients: \(a_{11}= -1\) \(a_{12}= -3\) \(a_{21}= 1\) \(a_{22}= -1\) So, the matrix \(A\) is: \(A = \begin{pmatrix} -1 & -3 \\ 1 & -1 \end{pmatrix}\)

Key concepts

Coefficient Matrix

In the context of linear systems of differential equations, the coefficient matrix plays a crucial role. A coefficient matrix, denoted as \( A \), captures all the numerical coefficients that link the system's individual equations. For the problem at hand, the task is to uncover this matrix from the given general solution. This involves:

• Identifying how each component of the vector \( \mathbf{y} \) relates to its derivative.
• The coefficients in \( A \) will describe these relationships.

In the exercise, we found the coefficient matrix:\[A = \begin{pmatrix} -1 & -3 \ 1 & -1 \end{pmatrix}\]This matrix effectively represents the linear relationships through which the system evolves over time.

General Solution

The general solution of a linear differential system describes how the system's variables change over time. In our case, the system is expressed through \( y_1(t) \) and \( y_2(t) \), each influenced by constants \( c_1 \) and \( c_2 \). These constants, combined with functions of time, depict general behaviors of the system:

• For \( y_1(t) \), the specific solution is:
  \( y_{1}(t) = c_{1} e^{-t}(1 + 2t) + 4c_{2} t e^{-t} \)
• For \( y_2(t) \), it is:
  \( y_{2}(t) = -c_{1} t e^{-t} + c_{2} e^{-t}(1 - 2t) \)

Combining these equations captures how each part of the system varies, offering a full depiction of the system's dynamics. This solution becomes a base from which we derive the coefficient matrix.

First Derivatives

Differentiation is central to understanding how variables in a dynamic system change. Here, we calculate the first derivatives \( y_1'(t) \) and \( y_2'(t) \) to capture the rate of change of each function:

• \( y_1'(t) \) involves taking the derivative of the function:
  \( y_1'(t) = \frac{d}{dt}(c_{1} e^{-t}(1 + 2t) + 4c_{2} t e^{-t}) \)
• \( y_2'(t) \) is similarly computed:
  \( y_2'(t) = \frac{d}{dt}(-c_{1} t e^{-t} + c_{2} e^{-t}(1 - 2t)) \)

These derivatives are then expressed in terms of \( y_1(t) \) and \( y_2(t) \). As a result, they form a critical part of the differential equations that make up our system, leading us to establish the coefficients in matrix \( A \).

Matrix Representation

Representing the system of linear differential equations as a matrix is key for simplifying and solving complex systems. This involves structuring our system equations into a matrix form, which ties derivatives to the original functions:

• The derivative vector \( \mathbf{y}' \) is written in terms of the product of a matrix \( A \) and the vector \( \mathbf{y} \).
• In this setup, \[\begin{pmatrix}y_1'(t) \y_2'(t)\end{pmatrix} = A \begin{pmatrix}y_1(t) \y_2(t)\end{pmatrix}\]

The elegance of matrix representation lies in its ability to link each part of the system compactly, allowing us to easily compute and analyze solutions to even complex differential equations.