Question

The \(t\)-interval of interest is \(-\infty

Short answer

Based on the given information, verify that $y_1(t) = e^{-t}$ and $y_2(t) = 2e^{1-t}$ are solutions of the second-order linear homogeneous differential equation $y^{\prime \prime}+2 y^{\prime}+y=0$. Calculate the Wronskian and determine if these two functions form a fundamental set of solutions. If yes, find the unique solution of the given initial value problem. Answer: The given functions $y_1(t)$ and $y_2(t)$ are solutions of the differential equation, but they do not form a fundamental set of solutions. Therefore, we cannot find the unique solution of the given initial value problem using these functions.

Step-by-step solution

Find the derivatives of \(y_1(t)\) and \(y_2(t)\)

First, we need to find the first and second derivatives of both given functions. For \(y_1(t) = e^{-t}\): $$y_1'(t) = -e^{-t}$$ $$y_1''(t) = e^{-t}$$ For \(y_2(t) = 2e^{1-t}\): $$y_2'(t) = -2e^{1-t}$$ $$y_2''(t) = 2e^{1-t}$$ Now we have the first and second derivatives of both given functions.

Verify if the given functions are solutions of the differential equation

To verify that \(y_1(t)\) and \(y_2(t)\) are solutions of the given differential equation: \(y^{\prime \prime}+2 y^{\prime}+y=0\) We need to substitute the derivatives that we found in Step 1 into the differential equation, and then check if it holds true. For \(y_1(t)\): $$y_1''(t) + 2 y_1'(t) + y_1(t) = e^{-t} - 2e^{-t} + e^{-t} = 0$$ For \(y_2(t)\): $$y_2''(t) + 2 y_2'(t) + y_2(t) = 2e^{1-t} + 4e^{1-t} + 2e^{1-t} = 0$$ As seen in both cases, the differential equation holds true, so both \(y_1(t)\) and \(y_2(t)\) are solutions of the given differential equation.

Calculate the Wronskian and verify if they form a fundamental set of solutions

The Wronskian is defined as: $$W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}$$ We substitute the derivatives that we found in Step 1: $$W(y_1, y_2) = \begin{vmatrix} e^{-t} & 2e^{1-t} \\ -e^{-t} & -2e^{1-t} \end{vmatrix} = -2e^{1-2t} - (-2e^{1-2t}) = 0$$ As the Wronskian, \(W(y_1, y_2) = 0\), the two functions do not form a fundamental set of solutions.

Conclusion

Since the given functions \(y_1(t)\) and \(y_2(t)\) do not form a fundamental set of solutions, we cannot find the unique solution of the given initial value problem using these functions.

Key concepts

Wronskian

The Wronskian is a critical concept in the study of differential equations, particularly when assessing the relationship between two functions that are purported to be solutions of a given differential equation. Named after the Polish mathematician Józef Hoene-Wronski, it's a form of determinant that can be used to establish the linear independence of a set of functions.

To calculate the Wronskian for two functions, let's denote these functions as f₁(x) and f₂(x), you would construct and evaluate the determinant of their first derivatives organized in a matrix format as follows:
\[ W(f_1, f_2) = \begin{vmatrix} f_1(x) & f_2(x) \ f_1'(x) & f_2'(x) \end{vmatrix} \]
This matrix contains the original functions in the first row and their derivatives in the second row. The Wronskian is zero if the functions are linearly dependent, which generally indicates that they do not form a fundamental set of solutions. However, a non-zero Wronskian signifies that the functions are linearly independent and typically form a fundamental set. In the context of the textbook problem, the Wronskian between the two proposed solutions was found to be zero, meaning they do not form a fundamental set of solutions.

Fundamental Set of Solutions

A fundamental set of solutions plays a pivotal role in the theory of linear differential equations, as it pertains to the solution space of these equations. For a differential equation of n-th order, a fundamental set consists of n linearly independent solutions. This set is crucial because any solution to the differential equation can be expressed as a linear combination of these foundational solutions.

The importance of a fundamental set arises from its ability to span the solution space. When we have a set of solutions that is fundamental, we can address a wide array of initial conditions and construct a general solution accordingly. The general solution will encompass all possible specific solutions for the given differential equation. In the exercise, determining whether the functions y₁(t) and y₂(t) form such a set was essential; yet, due to the Wronskian being zero, they fail to form this fundamental set.

Initial Value Problem

The initial value problem is a type of differential equation problem where the solution is not only required to satisfy the differential equation, but also to match certain specified conditions at the initial point--typically the value of the solution and its derivatives at that point. These conditions are known as initial values and strongly determine the unique solution to the differential equation.

Specifically, initial value problems are defined for ordinary differential equations and require a unique solution to exist and be continuous over the interval of interest. The conditions given at the initial point help to identify this unique trajectory among all possible solutions that satisfy the general equation. In the exercise provided, the initial value problem cannot be conclusively resolved with the functions given because they do not form a fundamental set of solutions, which is crucial for the existence of a unique solution.

Second Order Linear Differential Equation

A second order linear differential equation is a form of differential equation characterized by the highest derivative being the second derivative of the unknown function. The general form of this equation is:
\[ y'' + p(t)y' + q(t)y = g(t) \]
where p(t), q(t), and g(t) are given functions of t, and the solutions typically indicate how a certain quantity evolves over time.

The equation in our exercise is a homogenous second order linear differential equation, which implies that g(t) is zero, simplifying the equation to:
\[ y'' + 2y' + y = 0 \]
This specific form suggests that both the first and second derivatives are involved in determining the behavior of solutions. Such equations are indicative of processes with 'memory', where the rate of change not only depends on the current state but also on the velocity of change, as implied by the second derivative.