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

Short answer

#tag_title#Step 3: Equate corresponding elements and solve for A#tag_content# Equate the corresponding elements of the result of the matrix multiplication to that of the given second set of solutions: \(2t^2 - 1 = a_{11} + a_{21} \ln t + a_{31} t^2\) \(3 = a_{12} + a_{22} \ln t + a_{32} t^2\) \(\ln t^3 = a_{13} + a_{23} \ln t + a_{33} t^2\) Now, we can solve for the elements of matrix A. From the first equation: \(2t^2 - 1 = a_{11} + a_{21} \ln t + a_{31} t^2\) Comparing the constant, \(\ln t\), and \(t^2\) terms, we can deduce: \(a_{11} = -1\), \(a_{21} = 0\), \(a_{31} = 2\) From the second equation: \(3 = a_{12} + a_{22} \ln t + a_{32} t^2\) Comparing the constant, \(\ln t\), and \(t^2\) terms, we can deduce: \(a_{12} = 3\), \(a_{22} = 0\), \(a_{32} = 0\) From the third equation: \(\ln t^3 = a_{13} + a_{23} \ln t + a_{33} t^2\) Comparing the constant, \(\ln t\), and \(t^2\) terms, we can deduce: \(a_{13} = 0\), \(a_{23} = 3\), \(a_{33} = 0\) So, the matrix \(A\) is: \(A = \begin{bmatrix} -1 & 3 & 0\\ 0 & 0 & 3\\ 2 & 0 & 0 \end{bmatrix}\) #tag_title#Step 4: Determine whether the second set of solutions is fundamental#tag_content# To determine if the second set of solutions is also a fundamental set, we need to calculate the determinant of matrix \(A\): \(\det(A) = \begin{vmatrix} -1 & 3 & 0\\ 0 & 0 & 3\\ 2 & 0 & 0 \end{vmatrix} = 18\) Since the determinant of matrix \(A\) is non-zero, the second set of solutions \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}\) is also a fundamental set of solutions for the given differential equation. #Answer# The constant matrix A is: \(A = \begin{bmatrix} -1 & 3 & 0\\ 0 & 0 & 3\\ 2 & 0 & 0 \end{bmatrix}\) The second set of solutions \(\left\\{\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right\\}\) is also a fundamental set of solutions for the given differential equation since the determinant of matrix \(A\) is non-zero.

Step-by-step solution

Write down the given sets of solutions

We are given two sets of solutions: A fundamental set of solutions: \(y_{1}(t) = 1\), \(y_{2}(t) = \ln t\), \(y_{3}(t) = t^2\) Another set of solutions: \(\bar{y}_{1}(t) = 2t^2 - 1\), \(\bar{y}_{2}(t) = 3\), \(\bar{y}_{3}(t) = \ln t^3\)

Find the constant matrix A

We need to find a constant matrix \(A\) such that: \(\left[\bar{y}_{1}(t), \bar{y}_{2}(t), \bar{y}_{3}(t)\right] = \left[y_{1}(t), y_{2}(t), y_{3}(t)\right] A\) Let \(A = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}\) Then, we need to find the elements of matrix \(A\) such that: \(\begin{bmatrix} 2t^2 - 1 & 3 & \ln t^3 \end{bmatrix} = \begin{bmatrix} 1 & \ln t & t^2 \end{bmatrix} \times \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}\)

Key concepts

Fundamental Set of Solutions

A fundamental set of solutions in the context of differential equations refers to a set of solutions that can be used to express all possible solutions of a linear homogeneous differential equation. For a typical third-order differential equation, such as the one given in the exercise, the fundamental set consists of three functions. Each of these solutions is linearly independent of the others. This means no solution can be written as a combination of the others in the set.

In the exercise, the functions \( y_1(t) = 1 \), \( y_2(t) = \ln t \), and \( y_3(t) = t^2 \) form the fundamental set. This implies that any solution to the given differential equation can be expressed as a linear combination of these functions:

• \( c_1 y_1(t) + c_2 y_2(t) + c_3 y_3(t) \)

where \( c_1, c_2, \) and \( c_3 \) are constants which can be determined based on initial or boundary conditions. In essence, these solutions provide the complete "toolkit" needed to describe any solution to the differential equation.

Matrix Representation

Matrix representation offers a compact and powerful way to link different sets of solutions for differential equations. In this exercise, the idea is to determine a transformation matrix \( A \) that relates two sets of solutions. The first set is the fundamental set \( \{ y_1(t), y_2(t), y_3(t) \} \), and the second set is \( \{ \bar{y}_1(t), \bar{y}_2(t), \bar{y}_3(t) \} \).

The transformation matrix \( A \) is a \( 3 \times 3 \) matrix chosen such that:

• \[ \begin{bmatrix} \bar{y}_1(t) & \bar{y}_2(t) & \bar{y}_3(t) \end{bmatrix} = \begin{bmatrix} y_1(t) & y_2(t) & y_3(t) \end{bmatrix} \cdot A \]

The task involves finding the elements \( a_{ij} \) of the matrix \( A \), which act as the weights to transform the fundamental solutions into the new set. Solving this involves setting the products equal and forming a system of equations to solve for each \( a_{ij} \). This process unveils how the solutions of two sets are interrelated by linear transformations.

Determinant Calculation

Determinant calculation plays an essential role in identifying whether a set of solutions forms a fundamental set. For the matrix \( A \) found previously, calculating the determinant \( \det(A) \) helps determine if the set \( \{ \bar{y}_1(t), \bar{y}_2(t), \bar{y}_3(t) \} \) is also a fundamental set.

If the determinant \( \det(A) eq 0 \), it confirms that the second set is linearly independent and thus forms a fundamental set of solutions, much like the first. On the other hand, if \( \det(A) = 0 \), the solutions in \( A \) do not span the entire solution space independently, meaning they are not linearly independent.

• This step is crucial because the determinant gives insight into the structure and properties of the solution sets related through \( A \).

By checking the determinant, one can quickly verify whether transformation through \( A \) preserves these key properties of a fundamental set.