Question

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

Short answer

Based on the given information, we can conclude that: (a) Both functions \(y_1(t) = 2e^t\) and \(y_2(t) = e^{-t+3}\) are solutions to the differential equation \(y^{\prime\prime} - y = 0\). (b) The Wronskian of the two functions is calculated as \(W(y_1, y_2) = 0\). Therefore, the given functions do not form a fundamental set of solutions. (c) Since the functions do not form a fundamental set of solutions, we cannot use them to determine the unique solution of the initial value problem with the given initial conditions.

Step-by-step solution

Verify the solutions

To verify that both functions \(y_1(t)=2e^t\) and \(y_2(t)=e^{-t+3}\) are solutions to the differential equation \(y^{\prime\prime}-y=0\), we need to plug each function into the equation and check if the equation holds true. First, check \(y_1(t)=2e^t\): 1. Calculate the first derivative: \(y^{\prime}= \frac{d}{dt}(2e^t)=2e^t\) 2. Calculate the second derivative: \(y^{\prime\prime}= \frac{d^2}{dt^2}(2e^t)=2e^t\) 3. Plug derivatives into the equation: \(2e^t - 2e^t=0\) Since the equation is valid, \(y_1(t)\) is a solution. Now, check \(y_2(t)=e^{-t+3}\): 1. Calculate the first derivative: \(y^{\prime}= \frac{d}{dt}(e^{-t+3})=-e^{-t+3}\) 2. Calculate the second derivative: \(y^{\prime\prime}= \frac{d^2}{dt^2}(e^{-t+3})=e^{-t+3}\) 3. Plug derivatives into the equation: \(e^{-t+3} - e^{-t+3}=0\) Since the equation is valid, \(y_2(t)\) is also a solution.

Calculate the Wronskian

The Wronskian is defined as \(W(y_1, y_2) = y_1 y_2^{\prime} - y_2 y_1^{\prime}\). Calculating the Wronskian for the given solutions: \(W(y_1, y_2) = (2e^t)(-e^{-t+3}) - (e^{-t+3})(2e^t) = -2e^{3} + 2e^{3} = 0\) Since the Wronskian is zero, the two functions do NOT form a fundamental set of solutions.

Determine the unique solution (skipped)

As the Wronskian is zero, the given functions do not form a fundamental set of solutions. Thus, we can't use them to determine the unique solution of the initial value problem.

Key concepts

Fundamental Set of Solutions

A fundamental set of solutions is crucial when solving linear differential equations. It refers to a set of solutions from which all possible solutions of a differential equation can be formed. If two solutions of a second-order linear differential equation are linearly independent, they form this fundamental set. This means you can express any solution as a linear combination of these two solutions.
To determine whether two solutions form a fundamental set, we use the Wronskian, a determinant that gives us information about their linear independence. If the Wronskian is non-zero over an interval, the solutions are independent, and therefore, they form a fundamental set. Understanding this concept helps to solve the equation more effectively.

Wronskian

The Wronskian is named after the Polish mathematician Józef Hoene-Wroński. It is a determinant used in the analysis of differential equations to check whether a set of functions are linearly independent. For two functions, the Wronskian is calculated as:

\[ W(y_1, y_2) = \begin{vmatrix} y_1(t) & y_2(t) \y_1'(t) & y_2'(t) \\end{vmatrix} \]
This simplifies to:
\[ W(y_1, y_2) = y_1(t)y_2'(t) - y_2(t)y_1'(t) \]

If the Wronskian is zero for some interval, it suggests that the functions are linearly dependent on that interval. From the example, since the Wronskian is zero, the given solutions are not independent, and do not form a fundamental set. Thus, they cannot cover all possible solutions of the differential equation.

Initial Value Problem

An initial value problem (IVP) in differential equations specifies the value of the function and possibly its derivatives at a particular point. This contrasts with a boundary value problem, which specifies the values at the boundaries of the interval.
The IVP is important because it provides enough information to find a unique solution to a differential equation. Given a differential equation and an IVP, you can solve to find a specific function that fits both the equation and the initial conditions. For instance, conditions like \(y(-1) = 1\), and \(y'(-1) = 0\) are typical in formulating an IVP.

Solutions Verification

Verifying solutions is crucial before solving differential equations further. It involves substituting potential solutions back into the differential equation to ensure they satisfy it.
For instance, given a differential equation like \(y'' - y = 0\), and solutions \(y_1(t) = 2e^t\) and \(y_2(t) = e^{-t+3}\), we check each by differentiating according to the equation's requirements:

• Find the first and second derivatives of each solution.
• Substitute these into the original equation.
• Check if the resulting expression equals zero, verifying they satisfy the equation.

This process ensures that the candidates are valid solutions to the differential equation, which is foundational in accurately solving the problem.