Problem 17 - 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 \((-\infty, \infty)\). $$ \mathbf{X}_{1}=\left(\begin{array}{l} 1 \\ 1 \end{array}\right) e^{-2 t}, \mathbf{X}_{2}=\left(\begin{array}{r} 1 \\ -1 \end{array}\right) e^{-6 t} $$

Step-by-Step Solution

Answer

The vectors form a fundamental set of solutions on the interval \((-\infty, \infty)\).

Step 1: Understand the Problem

We need to determine if the given vectors \( \mathbf{X}_{1} \) and \( \mathbf{X}_{2} \) form a fundamental set of solutions to the differential equation \( \mathbf{X}^{\prime} = \mathbf{A X} \) on the interval \(( -\infty, \infty )\). To do this, we must check if these solutions are linearly independent over the interval.

Step 2: Calculate the Wronskian

For two solutions to be linearly independent, their Wronskian must be non-zero. The Wronskian \( W(\mathbf{X}_1, \mathbf{X}_2) \) is calculated as the determinant of the matrix formed by \( \mathbf{X}_1 \) and \( \mathbf{X}_2 \):\[ W(\mathbf{X}_1, \mathbf{X}_2) = \begin{vmatrix} e^{-2t} & e^{-6t} \ e^{-2t} & -e^{-6t} \end{vmatrix} \]Calculate this determinant.

Step 3: Compute the Determinant

Calculate the determinant from Step 2:\(W(\mathbf{X}_1, \mathbf{X}_2) = (e^{-2t})(-e^{-6t}) - (e^{-2t})(e^{-6t})\)\(= -e^{-8t} - e^{-8t} \)\(= -2e^{-8t} \).

Step 4: Evaluate the Wronskian

Since \(-2e^{-8t} eq 0\) for any \( t \in (-\infty, \infty) \), the Wronskian is non-zero across the entire interval.

Step 5: Conclusion

Because the Wronskian is non-zero for all \( t \) in the interval \( (-\infty, \infty) \), the vectors \( \mathbf{X}_{1} \) and \( \mathbf{X}_{2} \) are linearly independent and form a fundamental set of solutions for the given interval.

Key Concepts

Wronskian

The Wronskian is a very handy tool in the study of differential equations, especially when analyzing the solutions' independence. It's a determinant associated with a set of functions or vectors. You calculate the Wronskian by forming a square matrix from these functions and their derivatives. Then compute its determinant. The most common use of the Wronskian is to check for linear independence between solutions.

In our specific exercise, the Wronskian is used to assess whether the given vectors \( \mathbf{X}_1 \) and \( \mathbf{X}_2 \) are independent. The process involves

• Setting up a matrix using these vectors.
• Computing its determinant.

For our vectors, the calculation is:\[ W(\mathbf{X}_1, \mathbf{X}_2) = \begin{vmatrix} e^{-2t} & e^{-6t} \ e^{-2t} & -e^{-6t} \end{vmatrix} \]By working out the determinant, you end up with \(-2e^{-8t}\). What's crucial here is that as long as the Wronskian is not zero across the interval (from minus infinity to infinity in this case), it confirms linear independence.

linear independence

The concept of linear independence is fundamental in linear algebra and extends its usefulness into differential equations. A set of vectors or functions is said to be linearly independent if no vector in the set can be written as a combination of the others. In the context of differential equations, we talk about the linear independence of solutions.

How do we check for independence? One efficient method is using the Wronskian. When you compute the Wronskian of the solutions and find it to be non-zero, it indicates linear independence. Hence, in the exercise, since the Wronskian \(-2e^{-8t}\) is always non-zero, \( \mathbf{X}_1 \) and \( \mathbf{X}_2 \) are linearly independent solutions.

This independence is exactly what we want when establishing a fundamental set of solutions, as it ensures the solutions span the space of possible solutions for the differential equation in consideration.

fundamental set of solutions

In differential equations, a fundamental set of solutions is a set that encompasses all possible solutions of the system or equation being studied. For linear systems, finding a fundamental set of solutions is crucial because it allows us to represent any solution as a combination of the items in this set.

For a set of solutions to be considered fundamental, they must satisfy specific criteria:

• The solutions are linearly independent over the interval in question.
• The Wronskian is non-zero throughout the interval.

Having checked that \( \mathbf{X}_1 \) and \( \mathbf{X}_2 \) are linearly independent due to the non-zero Wronskian, we can confidently say that they form a fundamental set of solutions over the given interval. This means that every solution to the differential system \( \mathbf{X}' = \mathbf{A X} \) can be expressed through these two vectors. So, understanding and finding such a set not only verifies our initial problem but unlocks the full span of this system's solutions.