Question

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

Short answer

Answer: No, the given functions do not form a fundamental set of solutions since their Wronskian equals to zero.

Step-by-step solution

1. Verify Solutions

We are given the differential equation \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) and two functions \(y_{1}(t)=2 e^{t}\) and \(y_{2}(t)=e^{2 t}\). We need to show that both functions satisfy the given equation. First, let's find the first and second derivatives of both functions: \(y_{1}'(t)=2 e^{t}\) \(y_{1}''(t)=2 e^{t}\) \(y_{2}'(t)=2 e^{2 t}\) \(y_{2}''(t)=4 e^{2 t}\) Now, let's plug these first and second derivatives into the differential equation and see if they satisfy the equation: For \(y_1(t)\): \((2 e^{t})-(3\times 2 e^{t})+(2 \times 2 e^{t})=0\) Simplifying, we get: \(2e^{t}=2e^{t}\) So \(y_1(t)\) is indeed a solution. For \(y_2(t)\): \((4 e^{2 t})-(3\times 2 e^{2 t})+(2 \times e^{2 t})=0\) Simplifying, we get: \(4e^{2t}=4e^{2t}\) So \(y_2(t)\) is also a solution. Both functions are solutions of the differential equation.

2. Calculate the Wronskian

To check if the two functions form a fundamental set of solutions, we need to calculate their Wronskian and see if it's non-zero. The Wronskian of two functions, \(y_1(t)\) and \(y_2(t)\), is defined as: \(W(y_1(t),y_2(t))=y_1(t)y_2'(t)-y_1'(t)y_2(t)\) Therefore, \(W(2 e^{t}, e^{2 t})=(2 e^{t})(2 e^{2 t})-(2 e^{t})(e^{2 t})\) Simplifying, we get: \(W(2 e^{t}, e^{2 t})=2 e^{3 t}-2 e^{3 t}=0\) Since the Wronskian equals zero, the two functions do not form a fundamental set of solutions.

3. Determine Unique Solution

As we found out earlier, the given functions do not form a fundamental set of solutions and hence we cannot determine the unique solution of the initial value problem using these functions.

Key concepts

Wronskian

In differential equations, the Wronskian is a determinant used to identify whether a set of solutions is linearly independent. This is crucial because linear independence of functions ensures they can form a basis for the solution space of the differential equation. For two functions, the Wronskian is calculated as follows:

• Take the functions, say \(y_1\) and \(y_2\), and compute their derivatives: \(y_1'(t)\) and \(y_2'(t)\).
• Use the formula: \[ W(y_1(t), y_2(t)) = y_1(t)y_2'(t) - y_1'(t)y_2(t) \]

In the solution example, the functions \(y_1(t) = 2e^t\) and \(y_2(t) = e^{2t}\) were used. By substituting these and their derivatives into the Wronskian formula, it resulted in zero:\[ W(2 e^{t}, e^{2 t})=2 e^{3 t} - 2 e^{3 t} = 0 \]A zero Wronskian indicates that the functions are not linearly independent over the interval, thus they do not form a fundamental set of solutions.Understanding the Wronskian helps in analyzing the solution structure of differential equations in terms of dependency and uniqueness of solutions.

Fundamental Set of Solutions

A fundamental set of solutions is a collection of solutions to a linear differential equation that span the entire solution space. The most crucial aspect of having a fundamental set is that the solutions must be linearly independent. This ensures that any solution of the differential equation can be expressed as a linear combination of the set. The verification of a fundamental set involves several steps:

• Compute the associated Wronskian for the set of functions.
• Check the Wronskian value over the given interval.
• If the Wronskian is non-zero, the functions are linearly independent, forming a fundamental set.

In the discussed example, even though \(y_1(t) = 2e^t\) and \(y_2(t) = e^{2t}\) satisfy the original differential equation \(y'' - 3y' + 2y = 0\), their Wronskian is zero:\[ W(2e^{t}, e^{2t}) = 0 \]Therefore, they do not constitute a fundamental set of solutions and cannot span the solution space needed to solve any related initial value problem.

Initial Value Problem

An initial value problem involves finding a specific solution to a differential equation that satisfies certain initial conditions. Initial value problems are common in various fields such as physics and engineering, where initial states of a system are known and future states are to be determined. The process usually entails:

• Ensuring the underlying functions form a fundamental set of solutions to the given differential equation.
• Using the initial conditions to find the constants in the general solution.

For the exercise at hand, we needed to solve:\[y'' - 3y' + 2y = 0\]with initial conditions \(y(-1) = 1\) and \(y'(-1) = 0\). However, since the Wronskian was zero meaning \(y_1(t) = 2e^t\) and \(y_2(t) = e^{2t}\) are not a fundamental set, the solution using these functions could not be uniquely determined. It's critical to find another pair of linearly independent solutions to successfully solve the initial value problem for the specified conditions.