Problem 18 - Advanced Engineering Mathematics

Question

In Problems 17-20, the given vectors are solutions of a system \(\mathbf{X}^{\prime}=\mathbf{A X}\). Determine whether the vectors form a fundamental set on the interval \((-\infty, \infty)\). $$ \mathbf{X}_{1}=\left(\begin{array}{r} 1 \\ -1 \end{array}\right) e^{t}, \mathbf{X}_{2}=\left(\begin{array}{l} 2 \\ 6 \end{array}\right) e^{t}+\left(\begin{array}{r} 8 \\ -8 \end{array}\right) t e^{t} $$

Step-by-Step Solution

Answer

The vectors form a fundamental set on \((- fty, fty)\) because the Wronskian is nonzero.

Step 1: Identify Vectors Form

The problem provides two vector solutions of the system \( \mathbf{X}' = \mathbf{A} \mathbf{X} \). We have \( \mathbf{X}_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{t} \) and \( \mathbf{X}_2 = \begin{pmatrix} 2 \ 6 \end{pmatrix} e^{t} + \begin{pmatrix} 8 \ -8 \end{pmatrix} t e^{t} \). We'll determine if these form a fundamental set.

Step 2: Set Up the Wronskian

To check if the vectors \( \mathbf{X}_1 \) and \( \mathbf{X}_2 \) form a fundamental set, compute the Wronskian, which is given by the determinant:\[ W(t) = \begin{vmatrix} 1 e^{t} & (2 + 8t) e^{t} \ -1 e^{t} & (6 - 8t) e^{t} \end{vmatrix} \]

Step 3: Compute the Determinant of the Wronskian

Expand the determinant:\[W(t) = \begin{vmatrix} e^{t} & (2 + 8t)e^{t} \ -e^{t} & (6 - 8t)e^{t} \end{vmatrix} = e^{t}((6 - 8t)e^{t}) - ((-e^{t})(2 + 8t)e^{t})\]Simplifying gives:\[ W(t) = e^{2t}(6 - 8t + 2 + 8t) = 8 e^{2t} \]

Step 4: Analyze the Wronskian

The Wronskian \( W(t) = 8 e^{2t} \) is nonzero for \( t \in (-\infty, \infty) \) because \( e^{2t} eq 0 \) for all real \( t \). A nonzero Wronskian indicates that the vectors form a fundamental set over the interval.

Key Concepts

Wronskian determinant

The Wronskian determinant is a useful tool for checking if a set of functions are linearly independent on a given interval. When dealing with systems of differential equations, it also helps determine if the solutions form a fundamental set.
In our case, the vectors \(\mathbf{X}_1\) and \(\mathbf{X}_2\) need to be checked for linear independence to see if they form a fundamental set over \((-\infty, \infty)\). This is done by computing the Wronskian of the two functions:

• The Wronskian \(W(t)\) is defined as the determinant of the matrix formed by the vectors.
• If \(W(t)\) is non-zero for all \(t\) in the interval, the functions are linearly independent.

In our solution:

• The Wronskian calculated is \(W(t) = 8e^{2t}\).
• This determinant is non-zero for all \(t\), since exponential functions do not equal zero.
• Therefore, \(\mathbf{X}_1\) and \(\mathbf{X}_2\) are linearly independent and form a fundamental set on \((-\infty, \infty)\).

Linear independence

Linear independence is a concept that shows that one vector in a set cannot be made from combinations of the others. It is a crucial characteristic when determining if solutions to a system of differential equations form a fundamental set.
In simpler terms, if you have two vectors and you cannot create one vector by scaling the other or adding them together, they're linearly independent.
In mathematical terms:

• For vectors \(\mathbf{X}_1\) and \(\mathbf{X}_2\), they are linearly independent if when you create any linear combination equal to zero, the only solution is the trivial one where all coefficients equal zero.

The Wronskian determinant is a powerful method to check this quickly. When the Wronskian is non-zero, it indicates that the vectors are linearly independent over the interval of interest.
Since the Wronskian in the solution was non-zero, the vectors \(\mathbf{X}_1\) and \(\mathbf{X}_2\) are indeed independent, confirming they form a fundamental set.

Systems of differential equations

Systems of differential equations involve multiple equations that pertain to multiple functions. These systems describe various real-world phenomena, such as population dynamics, electrical circuits, and mechanical systems. The solutions to these systems represent how these phenomena evolve over time.
In our exercise, we analyze the system \(\mathbf{X}' = \mathbf{A} \mathbf{X}\) with solutions given as vectors. Solving these systems often involves finding a fundamental set of solutions that cover all possibilities for the system's behavior under given conditions.
Here's what this means:

• A fundamental set of solutions allows us to write any solution of the system as a linear combination of vectors in this set.
• This makes them extremely valuable in fully understanding a system's dynamics.
• In our problem, confirming the fundamental set ensures we can understand every possible behavior of the given differential equation.

Thus, the power of the fundamental set of solutions is in offering complete flexibility and understanding of the system’s solutions for every possible scenario that may arise.