Question

Determine whether the given functions form a fundamental set of solutions for the linear system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -1 & 2 \\ 0 & -1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}_{1}(t)=\left[\begin{array}{c} e^{-t} \\ 0 \end{array}\right], \quad \mathbf{y}_{2}(t)=\left[\begin{array}{l} e^{-t} \\ e^{-t} \end{array}\right] $$

Short answer

Answer: Yes, the given functions form a fundamental set of solutions for the linear system.

Step-by-step solution

Write down the given functions and system

We are given the following functions and linear system: $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -1 & 2 \\ 0 & -1 \end{array}\right] \mathbf{y}, \quad \mathbf{y}_{1}(t)=\left[\begin{array}{c} e^{-t} \\ 0 \end{array}\right], \quad \mathbf{y}_{2}(t)=\left[\begin{array}{l} e^{-t} \\ e^{-t} \end{array}\right] $$

Calculate the Wronskian

Now we need to calculate the Wronskian of the given functions \(\mathbf{y}_{1}(t)\) and \(\mathbf{y}_{2}(t)\). The Wronskian of two \(2\times1\) functions is defined as: $$ W(\mathbf{y}_{1},\mathbf{y}_{2})(t) = \begin{vmatrix} y_{1}(t) & y_{2}(t)\\ y_{1}'(t) & y_{2}'(t) \end{vmatrix} $$ Plug in \(\mathbf{y}_{1}(t)\) and \(\mathbf{y}_{2}(t)\) into the definition above: $$ W(\mathbf{y}_{1},\mathbf{y}_{2})(t) = \begin{vmatrix} e^{-t} & e^{-t}\\ 0 & -e^{-t} \end{vmatrix}. $$

Compute the determinant of the Wronskian

Now we compute the determinant of the Wronskian: $$ W(\mathbf{y}_{1},\mathbf{y}_{2})(t) = (e^{-t})(-e^{-t}) - (0)(e^{-t}) = -(e^{-2t}). $$

Determine if the Wronskian is nonzero for any t

We have obtained the Wronskian as: $$ W(\mathbf{y}_{1},\mathbf{y}_{2})(t) = -(e^{-2t}) $$ Considering the exponential function is positive for all real numbers, we can conclude that the Wronskian is nonzero for all \(t\). Therefore, the given functions \(\mathbf{y}_{1}(t)\) and \(\mathbf{y}_{2}(t)\) form a fundamental set of solutions for the linear system \(\mathbf{y}^{\prime} = A\mathbf{y}\).

Key concepts

Wronskian

The Wronskian is a valuable tool in the study of differential equations, particularly when assessing the linear independence of solutions. It is named after the Polish mathematician Józef Maria Hoene-Wroński.

The Wronskian for a pair of functions, say \( f(t) \) and \( g(t) \), is a determinant used to determine whether the set of functions is linearly independent. If the Wronskian is nonzero for any point in the interval of interest, the functions are linearly independent. On the contrary, if the Wronskian is identically zero on the interval, the functions are said to be linearly dependent.

For the set of solutions in a linear differential equation, the existence of a non-zero Wronskian at any point implies that the solutions form a fundamental set, which is essential in forming the general solution of the system. Applying this concept to the problem at hand, we calculated the Wronskian as \( -(e^{-2t}) \), which is never zero. Thus, we affirm that the functions provided form a fundamental set of solutions.

Linear Systems of Differential Equations

Linear systems of differential equations consist of multiple linear differential equations involving several unknown functions. Such a system is described by an equation of the form \( \mathbf{y}' = A\mathbf{y} \) for a matrix \( A \) and a vector function \( \mathbf{y} \).

In these systems, the goal is often to find a solution for \( \mathbf{y} \) that satisfies the equation over an interval. A fundamental set of solutions for such a system provides the basis that can be used to express any possible solution of the system, which is akin to finding a basis in vector space that can span the entire space.

Understanding and solving these systems often involve utilizing concepts such as eigenvalues, eigenvectors, and matrix exponentials, as well as calculating determinants and Wronskians, which we've seen in action in this exercise.

Determinant Calculation

Determinants are mathematical objects that are particularly useful in linear algebra, analysis, and related areas. The determinant of a square matrix provides crucial insights into the matrix's properties, such as whether it is invertible or singular.

The calculation involves a specific arithmetic expression of the elements of the matrix. For a 2x2 matrix \( A \) with entries \( a, b, c, d \) as \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) the determinant \( |A| \) is calculated as \( ad - bc \). This formula extends, with more complexity, to larger square matrices.

In the context of our exercise, the determinant was computed for the Wronskian matrix, which played a pivotal role in determining the linear independence of solutions to the differential system.

Exponential Functions

Exponential functions are ubiquitous in mathematics and can be identified by their base raised to a variable exponent, typically expressed as \( f(x) = b^x \) with \( b > 0 \) and \( b eq 1 \). These functions are important due to their unique properties and behaviors, including the fact that they are their own derivatives and integrals up to a constant (for the natural exponential function \( e^x \)).

They appear in numerous applications across different fields like physics, finance, and biology, particularly in modeling growth and decay processes. In the exercise, exponential functions provided by \( \mathbf{y}_1(t) \) and \( \mathbf{y}_2(t) \) feature markedly in the solutions for the differential system and influence the calculation of the Wronskian, reinforcing their significance.