Problem 1 - Differential Equations and Linear Algebra

Question

Verify that \(\mathbf{x}_{1}(t)=\left[\begin{array}{l}e^{t^{2}-t} \\\ -1\end{array}\right]\) and \(\mathbf{x}_{2}(t)=\left[\begin{array}{c}0 \\ 2 e^{t}\end{array}\right]\) are linearly independent solutions to \(\mathbf{x}^{\prime}=A \mathbf{x},\) where $$ A=\left[\begin{array}{ll} 2 t-1 & 0 \\ e^{t-t^{2}} & 1 \end{array}\right] $$ Write the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\)

Step-by-Step Solution

Answer

The general solution to the given system \(\mathbf{x}' = A\mathbf{x}\) is \(\mathbf{x}(t) = \left[\begin{array}{l} c_{1}e^{t^{2}-t} \\ -c_{1}+2c_{2}e^{t} \end{array}\right]\), where \(c_{1}\) and \(c_{2}\) are arbitrary constants, and both \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are linearly independent solutions.

Step 1: Check whether \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are solutions to the system

: First, let's find the derivative of each vector with respect to their variable \(t\) as \(\mathbf{x}'=A\mathbf{x}\) needs to hold for both vectors. \(\mathbf{x}_{1}(t) = \left[\begin{array}{l}e^{t^{2}-t} \\-1\end{array}\right] \implies \mathbf{x}_{1}'(t) = \left[\begin{array}{l} (2t-1)e^{t^{2}-t}\\ 0\end{array}\right]\) \(\mathbf{x}_{2}(t) = \left[\begin{array}{l}0 \\ 2e^{t}\end{array}\right] \implies \mathbf{x}_{2}'(t) = \left[\begin{array}{l} 0 \\2e^{t}\end{array}\right]\) Now, we need to check if \(\mathbf{x}_{1}'(t) = A \mathbf{x}_{1}(t) = \left[\begin{array}{ll}2t-1 & 0 \\ e^{t-t^{2}} & 1\end{array}\right] \left[\begin{array}{l}e^{t^{2}-t}\\-1\end{array}\right] = \left[\begin{array}{l}(2t-1)e^{t^{2}-t}\\-e^{t^{2}-t}+e^{t-t^{2}}\end{array}\right]\) As \(-e^{t^{2}-t}+e^{t-t^{2}}=0\), we can conclude that \(\mathbf{x}_{1}(t)\) is a solution to the system. Next, we need to check if \(\mathbf{x}_{2}'(t) = A \mathbf{x}_{2}(t) = \left[\begin{array}{ll} 2t-1 & 0 \\ e^{t-t^{2}} & 1 \end{array}\right] \left[\begin{array}{l} 0 \\ 2e^{t} \end{array}\right] = \left[\begin{array}{l} 0 \\ 2e^{t}\end{array}\right]\) As the given equalities hold, we can conclude that both \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are solutions to the system.

Step 2: Show that \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are linearly independent

: To verify that the vectors \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are linearly independent, we need to show that for any real values \(c_{1}\) and \(c_{2}\), the following equation holds: \(c_{1}\mathbf{x}_{1}(t) + c_{2}\mathbf{x}_{2}(t) = \mathbf{0}\) if and only if: \(c_{1} = 0\) and \(c_{2} = 0\). \(c_{1}\mathbf{x}_{1}(t) + c_{2}\mathbf{x}_{2}(t) = c_{1}\left[\begin{array}{l}e^{t^{2}-t}\\-1\end{array}\right] + c_{2}\left[\begin{array}{l}0\\ 2e^{t}\end{array}\right] = \left[\begin{array}{l}c_{1}e^{t^{2}-t}\\-c_{1}+2c_{2}e^{t}\end{array}\right]\) This equation equals the zero vector \(\mathbf{0}\) if and only if \(c_{1}e^{t^{2}-t} = 0\) and \(-c_{1}+2c_{2}e^{t} = 0\). Now, the only way for the first component to be zero is if \(c_{1} = 0\). Since \(c_{1} = 0\), the second component becomes \(2c_{2}e^{t} = 0\). Since \(e^{t}\) is never zero, this implies that \(c_{2} = 0\). Thus, \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are linearly independent.

Step 3: Write the general solution to the system

: Since both \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are linearly independent solutions to the system, we can now write the general solution as: \(\mathbf{x}(t) = c_{1}\mathbf{x}_{1}(t) + c_{2}\mathbf{x}_{2}(t) = c_{1}\left[\begin{array}{l}e^{t^{2}-t}\\-1\end{array}\right] + c_{2}\left[\begin{array}{l}0\\ 2e^{t}\end{array}\right] =\left[\begin{array}{l} c_{1}e^{t^{2}-t} \\ -c_{1}+2c_{2}e^{t} \end{array}\right]\) where \(c_{1}\) and \(c_{2}\) are arbitrary constants.

Key Concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They play a crucial role in mathematics, as well as in the sciences and engineering, because they describe how things change. In this exercise, the differential equation given is a system \( \mathbf{x}' = A \mathbf{x} \), where \( A \) is a matrix and \( \mathbf{x} \) is a vector function of \( t \). The goal of solving a differential equation is to find the function \( \mathbf{x}(t) \) that satisfies the equation.

In general, solving differential equations involves:

• Finding the derivative of the function, which relates to the concept of rate of change.
• Substituting it back into the equation to check if it is satisfied.

Here, we found the derivative of \( \mathbf{x}_1(t) \) and \( \mathbf{x}_2(t) \), and verified that each satisfied the differential equation when substituted back. Thus, both functions are indeed solutions to the system of differential equations.

General Solution

The general solution to a differential equation encompasses all possible solutions. For linear systems like the one in this exercise, the general solution is often a combination of several linearly independent solutions. In this case, since \( \mathbf{x}_1(t) \) and \( \mathbf{x}_2(t) \) are solutions to the system and are linearly independent, they form the basis for the general solution.

The general solution to the system \( \mathbf{x}' = A \mathbf{x} \) can be written as:\[\mathbf{x}(t) = c_{1}\mathbf{x}_1(t) + c_{2}\mathbf{x}_2(t)\]where \( c_{1} \) and \( c_{2} \) are arbitrary constants determined by initial conditions or further constraints.

This approach is powerful because it allows us to express an infinite number of solutions (since \( c_{1} \) and \( c_{2} \) can take any real number values) that satisfy the original differential equation.

System of Equations

A system of equations involves multiple equations that are solved simultaneously. Each equation may involve common variables. In this problem, the system of differential equations is described by the matrix equation:
\( \mathbf{x}' = A\mathbf{x} \).
This type of system is particularly common in fields that use mathematical modeling to describe dynamic systems, like physics and engineering.

Solving a system of equations typically involves:

• Finding a set of solutions, one for each equation, such that all equations in the system are satisfied.
• Using these solutions to analyze the dynamics of the system.

In our case, the solutions \( \mathbf{x}_1(t) \) and \( \mathbf{x}_2(t) \) are calculated and then used to form a complete description of the general behavior of the system via the general solution.

Linearly Independent Vectors

Linearly independent vectors are vectors that cannot be expressed as a linear combination of each other. This concept is critical in writing general solutions for differential equations. If vectors are linearly independent, it means they span a space without redundancy.

In this exercise, we showed \( \mathbf{x}_1(t) \) and \( \mathbf{x}_2(t) \) are linearly independent by verifying the condition:

• If \( c_{1}\mathbf{x}_1(t) + c_{2}\mathbf{x}_2(t) = \mathbf{0} \), then \( c_{1} = 0 \) and \( c_{2} = 0 \).

This ensures that these vectors provide a complete basis for the solution space of the system of equations. They allow us to express any solution vector in terms of these basis vectors, providing flexibility to cover all potential scenarios described by the differential equation.