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 \(\operatorname{det}(A)\). \(y^{\prime \prime \prime}-y^{\prime \prime}=0\), \(\left\\{y_{1}(t), y_{2}(t), y_{3}(t)\right\\}=\left\\{1, t, e^{-t}\right\\}\). \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}=\left\\{1-2 t, t+2, e^{-(t+2)}\right\\}\)

Short answer

Question: Determine if the set \(\left\\{\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}\right\\}\) is a fundamental set of solutions. Answer: Since the determinant of matrix A, which connects the two sets of solutions, is not equal to 0 (\(\operatorname{det}(A) = e^{-2}\)), the set \(\left\\{\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}\right\\}\) is also a fundamental set of solutions.

Step-by-step solution

Find the matrix A

Let us represent the given solutions by their sets \(\left\\{y_{1}(t), y_{2}(t), y_{3}(t)\right\\}=\left\\{1, t, e^{-t}\right\\}\) and \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}=\left\\{1-2t, t+2, e^{-(t+2)}\right\\}\). We are looking for a constant 3x3 matrix \(A\) such that $$ \left[\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right]=\left[y_{1}(t), y_{2}(t), y_{3}(t)\right]A$$ Let the matrix A be $$ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$ Now we have to find the coefficients \(a_{ij}\) $$ \begin{bmatrix}1 - 2t, & t + 2, & e^{-(t+2)}\end{bmatrix} = \begin{bmatrix}1, & t, & e^{-t}\end{bmatrix}\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$ By comparing elements: \( 1 - 2t = a_{11} + a_{21}t + a_{31}e^{-t}\) \(t + 2 = a_{12} + a_{22}t + a_{32}e^{-t}\) \( e^{-(t+2)} = a_{13} + a_{23}t + a_{33}e^{-t}\) Considering t = 0, we get: \(a_{11} + a_{31} = 1\), \(a_{12} + a_{32} = 2\), \(a_{13} + a_{33} = e^{-2}\) Now we differentiate above equations and set t = 0: \(d(1-2t)/dt = -2 = a_{21} + a_{31}(-1)\) \(d(t + 2)/dt = 1 = a_{22} + a_{32}(-1)\) \(d(e^{-(t+2)})/dt = -e^{-(t+2)} = a_{23} + a_{33}(-1)\) set t = 0 and we have \(-e^{-2} = a_{23} - a_{33}\) By solving the above linear system of equations, we obtain: \(a_{11} = 1, a_{21} = -2, a_{31} = 0\) \(a_{12} = 0, a_{22} = 1, a_{32} = -1\) \(a_{13} = 0, a_{23} = e^{-2}, a_{33} = 0\) So, the matrix A is: $$ A = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & e^{-2} \\ 0 & -1 & 0 \end{bmatrix} $$

Determine if the second set is also a fundamental set

In order to determine if the set \(\left\\{\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}\right\\}\) is also a fundamental set, we need to calculate the determinant of matrix A: $$ \operatorname{det}(A) = \begin{vmatrix} 1 & 0 & 0 \\ -2 & 1 & e^{-2} \\ 0 & -1 & 0 \end{vmatrix} = (1)\begin{vmatrix} 1 & e^{-2} \\ -1 & 0 \end{vmatrix} $$ \(\operatorname{det}(A) = \left(1\right)\left(1\cdot0 - (-1) \cdot e^{-2}\right) = e^{-2}\) Since \(\operatorname{det}(A) \neq 0\), the set \(\left\\{\bar{y}_{1}, \bar{y}_{2}, \bar{y}_{3}\right\\}\) is also a fundamental set of solutions.

Key concepts

Fundamental Set of Solutions

Understanding the concept of a fundamental set of solutions is crucial when dealing with differential equations. In the realm of linear differential equations, a fundamental set of solutions is a group of solutions to the differential equation that are linearly independent. This means that no solution in the set can be written as a linear combination of the others. When dealing with higher-order differential equations, finding a fundamental set of solutions allows us to express the general solution of the equation as a linear combination of these basic solutions.

For example, consider the third-order differential equation given in the exercise: \(y''' - y'' = 0\). The provided set of functions \(\{1, t, e^{-t}\}\) is indeed a fundamental set of solutions for this equation. These functions are linearly independent over the interval for which the differential equation is defined, meaning they satisfy the requisite condition for being a fundamental set. This property is what allows us to use the superposition principle to build the general solution from these basic building blocks.

Constant Coefficient Matrix

The concept of a constant coefficient matrix is frequently applied when examining systems of linear differential equations. In the context of the problem, we are tasked with finding a constant 3x3 coefficient matrix \(A\) that relates two sets of solutions through a matrix multiplication. This matrix is 'constant' because its entries are not functions of time or any other variable; they are fixed numbers.

The equation \(\left[\bar{y}_1(t), \bar{y}_2(t), \bar{y}_3(t)\right] = \left[y_1(t), y_2(t), y_3(t)\right]A\) essentially states that each of the original functions (the fundamental set) can be transformed into the new set of solutions by applying this constant matrix. The matrix captures the linear transformation necessary to go from one basis of solutions to another. In solving for the matrix \(A\), we assumed the simple case where the coefficients \(a_{ij}\) are constants, greatly simplifying the process of determination.

Determinant Calculation

The determinant of a matrix is a scalar value that provides important information about the matrix and the linear transformation it represents. When determining if a set of solutions is also a fundamental set, the calculation of the determinant plays a pivotal role. If the determinant of matrix \(A\), obtained from the relationship between two sets of solutions, is non-zero, it confirms that the second set of solutions is also a fundamental set.

In the given exercise, we calculated the determinant of matrix \(A\) and found it to be \(e^{-2}\), which is not equal to zero. Hence, it signifies that the set \(\{\bar{y}_1, \bar{y}_2, \bar{y}_3\}\) is indeed a fundamental set. Determinants are also essential for solving systems of equations, evaluating eigenvalues, and understanding geometric properties of linear transformations, such as area or volume scaling and orientation.