Question

In each exercise, \(\left\\{y_{1}, y_{2}, y_{3}\right\\}\) is a fundamental set of solutions and \(\left\\{\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}\right\\}\) is a set of solutions. (a) Find a \((3 \times 3)\) constant matrix \(A\) such that \(\left[\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right]=\left[y_{1}(t), y_{2}(t), y_{3}(t)\right] A\). (b) Determine whether \(\left\\{\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}\right\\}\) is also a fundamental set by calculating \(\operatorname{det}(A)\). \(y^{\prime \prime \prime}-y^{\prime}=0\), \(\left\\{y_{1}(t), y_{2}(t), y_{3}(t)\right\\}=\left\\{1, e^{t}, e^{-t}\right\\}\), \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}=\\{\cosh t, 1-\sinh t, 2+\sinh t\\}\)

Short answer

Answer: No, the given set of solutions is not a fundamental set.

Step-by-step solution

Understand the given information

We are given the following: - The differential equation: \(y^{\prime \prime \prime}-y^{\prime}=0\) - The fundamental set of solutions: \(\left\\{y_{1}(t), y_{2}(t), y_{3}(t)\right\\}=\left\\{1, e^{t}, e^{-t}\right\\}\) - The set of solutions: \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}=\left\\{\cosh t, 1-\sinh t, 2+\sinh t\right\\}\)

Find the constant matrix A

We need to find a \((3 \times 3)\) constant matrix \(A\) such that: $ \left[\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right]=\left[y_{1}(t), y_{2}(t), y_{3}(t)\right] A. $ It can be written as: $ \begin{bmatrix} \cosh t & 1-\sinh t & 2+\sinh t \end{bmatrix} = \begin{bmatrix} 1 & e^{t} & e^{-t} \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}. $ Observe that we can write the given functions in the following way: \(\cosh t = \frac{e^{t} + e^{-t}}{2},\ -\) \(\sinh t = \frac{e^{t} - e^{-t}}{2}.\) We can express the set of solutions \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}\) using the sets \(\left\\{y_{1}(t), y_{2}(t), y_{3}(t)\right\\}\) as follows: $ \cosh t = \frac{1}{2}y_{1}(t) + \frac{1}{2}y_{2}(t) + \frac{1}{2}y_{3}(t) $ $ 1-\sinh t = y_{1}(t) - \frac{1}{2}y_{2}(t) + \frac{1}{2}y_{3}(t) $ $ 2+\sinh t = y_{1}(t) + \frac{1}{2}y_{2}(t) - \frac{1}{2}y_{3}(t) $ Thus, our matrix A is: $ A=\begin{bmatrix} 0 & 1 & 1 \\ 1/2 & -1/2 & 1/2 \\ 1/2 & 1/2 & -1/2 \end{bmatrix}. $

Calculate the determinant of A

To determine if the given set of solutions is also a fundamental set, we need to calculate the determinant of matrix A: $ \operatorname{det}(A)= \begin{vmatrix} 0 & 1 & 1 \\ 1/2 & -1/2 & 1/2 \\ 1/2 & 1/2 & -1/2 \end{vmatrix}. $ Expanding this determinant, we get: $ \operatorname{det}(A)=-1\begin{vmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{vmatrix}+1\begin{vmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{vmatrix}. $ Calculating the determinant: $ \operatorname{det}(A)=(-1)\left(\frac{1}{4} - \frac{1}{4}\right)+(1)\left(\frac{1}{4} - \frac{1}{4}\right)=0. $

Determine if the given set of solutions is a fundamental set

Since the determinant of matrix A is zero, we can conclude that the given set of solutions \(\left\\{\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}\right\\}\) is not a fundamental set.

Key concepts

Fundamental Set of Solutions

Understanding the concept of a 'fundamental set of solutions' is key to solving differential equations. When dealing with a linear homogeneous differential equation, a fundamental set of solutions is a group of functions that not only are solutions to the equation but also possess a specific property: they are linearly independent. This means that none of the solutions in the set can be written as a combination of the others.

Why is this important? Because linear independence ensures that the fundamental set can be used to construct any possible solution to the equation through linear combinations. For the case of a third-order differential equation, like the one in our exercise, we need three linearly independent solutions to make up a fundamental set. In our exercise, the set \(\{y_{1}(t), y_{2}(t), y_{3}(t)\}=\{1, e^{t}, e^{-t}\}\) was a fundamental set because the functions meet these criteria: they are solutions to the given differential equation and are linearly independent.

Constant Matrix

When working with sets of solutions to differential equations, we often encounter the concept of a 'constant matrix.' A constant matrix, as the name implies, is a matrix where each entry is a constant value; that is, the entries do not vary with the independent variable, which is often time (t) in differential equations.

In the context of our exercise, finding a constant matrix that transforms one set of solutions to another is a way of understanding the relationship between different sets of solutions. If such a matrix exists and is not singular (meaning its determinant is not zero), it can be used to validate the completeness and independence of the second set of solutions. In our example, we successfully found the constant matrix \(A\) that relates the fundamental set with another set of solutions \(\bar{y}_i(t)\), further aiding us in analyzing the properties of the latter.

Determinant of a Matrix

The determinant of a matrix is a special number that can provide a lot of information about the matrix itself. For square matrices, which have the same number of rows as columns, the determinant can signify whether the matrix is invertible, with non-zero determinants indicating an invertible matrix, and a determinant of zero signifying a non-invertible or singular matrix.

Why does this matter in our exercise? The determinant of matrix \(A\), which was found in the previous steps, gives insight into whether our secondary set of solutions \(\bar{y}_i(t)\) is also a fundamental set. Since the determinant was calculated to be zero, it tells us that matrix \(A\) is singular, hence the solutions \(\bar{y}_i(t)\) cannot be linearly independent and so do not form a fundamental set. This concept is crucial for further mathematical theory and applications, such as solving systems of linear equations and understanding the behavior of complex systems through their modeled equations.