Problem 4 - Differential Equations and Linear Algebra

Question

Show that the given functions are solutions of the system \(\mathbf{x}^{\prime}(t)=A(x) \mathbf{x}(t)\) for the given matrix \(A,\) and hence, find the general solution to the system (remember to check linear independence). If auxiliary conditions are given, find the particular solution that satisfies these conditions. $$\mathbf{x}_{1}(t)=\left[\begin{array}{c} e^{2 t} \\ -e^{2 t} \end{array}\right], \quad \mathbf{x}_{2}(t)=\left[\begin{array}{c} e^{2 t}(1+t) \\ -t e^{2 t} \end{array}\right]$$. $$A=\left[\begin{array}{rr} 3 & 1 \\\ -1 & 1 \end{array}\right]$$.

Step-by-Step Solution

Answer

The general solution of the system is: \(\mathbf{x}(t) = C_1 \left[\begin{array}{c} e^{2t} \\ -e^{2t} \end{array}\right] + C_2 \left[\begin{array}{c} e^{2t}(1+t) \\ -te^{2t} \end{array}\right]\), where the functions \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are both solutions to the system and are linearly independent.

Step 1: Verify the given functions as solutions

First, let's find the derivatives of the given functions: \(\mathbf{x}_1'(t) = \left[\begin{array}{c} 2e^{2t} \\ -2e^{2t} \end{array}\right]\) \(\mathbf{x}_2'(t) = \left[\begin{array}{c} 2e^{2t}(1+t) + e^{2t} \\ -2te^{2t} - e^{2t} \end{array}\right] = \left[\begin{array}{c} 3e^{2t}(1+t) \\ -e^{2t}(1+2t) \end{array}\right]\) Now we verify whether they satisfy the given system \(\mathbf{x}'(t) = A(x)\mathbf{x}(t)\): \(\mathbf{x}_1'(t) = A(x)\mathbf{x}_1(t) = \left[\begin{array}{rr} 3 & 1 \\ -1 & 1 \end{array}\right] \left[\begin{array}{c} e^{2 t} \\ -e^{2 t} \end{array}\right] = \left[\begin{array}{c} 2e^{2t} \\ -2e^{2t} \end{array}\right]\) \(\mathbf{x}_2'(t) = A(x)\mathbf{x}_2(t) = \left[\begin{array}{rr} 3 & 1 \\ -1 & 1 \end{array}\right] \left[\begin{array}{c} e^{2t}(1+t) \\ -te^{2t} \end{array}\right] = \left[\begin{array}{c} 3e^{2t}(1+t) \\ -e^{2t}(1+2t) \end{array}\right]\) As we've checked that \(\mathbf{x}_1'(t)\) and \(\mathbf{x}_2'(t)\) satisfy the given system, we can now check for linear independence.

Step 2: Check for linear independence

To check for linear independence, we can form the determinant of the Wronskian matrix: \(W(\mathbf{x}_1(t), \mathbf{x}_2(t)) = \det \left(\begin{array}{cc} e^{2t} & e^{2t}(1+t) \\ -e^{2t} & -te^{2t} \end{array}\right) = (-t e^{4t}) - ((1+t) e^{4t}) = -e^{4t}\) As the determinant of the Wronskian matrix is non-zero, the given functions are linearly independent.

Step 3: Find the general solution

Now that we have verified that the given functions are solutions to the system and they are linearly independent, we can find the general solution of the system: \(\mathbf{x}(t) = C_1 \mathbf{x}_1(t) + C_2 \mathbf{x}_2(t) = C_1 \left[\begin{array}{c} e^{2t} \\ -e^{2t} \end{array}\right] + C_2 \left[\begin{array}{c} e^{2t}(1+t) \\ -te^{2t} \end{array}\right]\)

Step 4: Find the particular solution (if auxiliary conditions are given)

If auxiliary conditions are given, we would need to solve for the constants \(C_1\) and \(C_2\) to find the particular solution, but we have not been given auxiliary conditions in this example. The general solution of the system is: \(\mathbf{x}(t) = C_1 \left[\begin{array}{c} e^{2t} \\ -e^{2t} \end{array}\right] + C_2 \left[\begin{array}{c} e^{2t}(1+t) \\ -te^{2t} \end{array}\right]\)

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that deals with vector spaces, matrices, and linear transformations. It provides the necessary tools to study linear equations, which are equations of the first degree, meaning they have variables that are raised to the power of one. In the context of differential equations, linear algebra is employed to solve systems of equations where the unknowns are functions rather than simple numbers.

Key concepts in linear algebra include:

• Vectors: Objects that have both a magnitude and a direction. In systems of differential equations, functions like \( \mathbf{x}_1(t) \) and \( \mathbf{x}_2(t) \) serve as vector solutions.
• Matrices: Rectangular arrays of numbers that can represent linear transformations. The matrix \( A \) is used to transform the vector function \( \mathbf{x}(t) \).
• Determinants: A scalar value derived from a square matrix that helps in determining linear independence.

By using linear algebra, we solve complex problems involving multiple equations efficiently, especially in the study of differential equations.

System of Differential Equations

A system of differential equations is a collection of multiple equations that involve more than one dependent variable and their derivatives. These systems are vital in modeling situations where multiple interacting quantities change over time, such as population dynamics or electrical circuits.

In our specific exercise, the system is given by:\[ \mathbf{x}'(t) = A \mathbf{x}(t) \]Here, \( \mathbf{x}(t) \) represents the set of unknown functions, and \( A \) is the matrix that defines how these functions are interrelated.

Solving such systems can involve verifying if proposed functions satisfy the original equations and whether these solutions are unique. The solutions \( \mathbf{x}_1(t) \) and \( \mathbf{x}_2(t) \) both satisfy the equation when plugged into the system. Thus, they act as particular solutions to the differential equation system.

Linear Independence

Linear independence is a crucial concept to determine if a set of functions provides a complete and non-redundant basis for solutions. For functions \( \mathbf{x}_1(t) \) and \( \mathbf{x}_2(t) \), checking their linear independence ensures that they form a basis for generating all possible solutions to the system and are not simply scaled versions of one another.

To check linear independence, we calculate the Wronskian determinant from the solutions. If the Wronskian is non-zero, the functions are linearly independent. For this exercise, we calculate:\[ W(\mathbf{x}_1(t), \mathbf{x}_2(t)) = (-t e^{4t}) - ((1+t) e^{4t}) = -e^{4t} \]Since \( -e^{4t} eq 0 \), the functions are linearly independent. This guarantees that we can use them to form the general solution by linear combinations.

General Solution

The general solution of a system of differential equations encompasses all possible solutions, formed by combining linearly independent solutions with arbitrary constants. Once linear independence of solutions is established, we can express all solutions as:\[ \mathbf{x}(t) = C_1 \mathbf{x}_1(t) + C_2 \mathbf{x}_2(t) \]where \( C_1 \) and \( C_2 \) are constants.

This combination accounts for any initial conditions or specific details provided later. In modeling real-world systems, these constants can be adjusted to fit specific scenarios when initial or boundary conditions are known.

The general solution encodes the system's dynamics and flexibility, allowing for modifications as needed based on external parameters and conditions, making it a powerful tool in analyzing differential equations.