Question

Consider the given differential equation on the interval \(-\infty

Short answer

Answer: Yes, the solutions form a fundamental set.

Step-by-step solution

Write down the given differential equation and initial conditions

The given third-order differential equation is $$ y^{\prime \prime \prime}+e^t y^{\prime \prime}+y=0. $$ The initial conditions are: $$ y_1(1)=0, \quad y_1^{\prime}(1)=1, \quad y_1^{\prime \prime}(1)=1, $$ $$ y_2(1)=1, \quad y_2^{\prime}(1)=-1, \quad y_2^{\prime \prime}(1)=0, $$ $$ y_3(1)=-1, \quad y_3^{\prime}(1)=0, \quad y_3^{\prime \prime}(1)=0. $$

Set up the Wronskian determinant

The Wronskian determinant for a set of 3 functions is given by: $$ W(y_1, y_2, y_3) = \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1^\prime & y_2^\prime & y_3^\prime \\ y_1^{\prime\prime} & y_2^{\prime\prime} & y_3^{\prime\prime} \end{vmatrix}. $$

Calculate the Wronskian using the given initial conditions

Substitute the initial conditions into the Wronskian determinant: $$ W(y_1, y_2, y_3)\big|_{t=1} = \begin{vmatrix} 0 & 1 & -1 \\ 1 & -1 & 0 \\ 1 & 0 & 0 \end{vmatrix}. $$ Compute the determinant using cofactor expansion: $$ W(y_1, y_2, y_3)\big|_{t=1} = 1(-1 - 0) + 1(0 - 1) + (-1)(-1 - 0) = -1 -1 + 1 = -1. $$

Determine if the solutions form a fundamental set

Since the Wronskian is not equal to zero at t=1, the solutions \(y_1\), \(y_2\), and \(y_3\) form a linearly independent set, which indicates that they form a fundamental set.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They express relationships between these functions and are used to model a wide array of real-world phenomena, such as motion, heat, electricity, and more.

In the given exercise, we work with a third-order differential equation: \( y^{\prime \prime \prime} + e^t y^{\prime \prime} + y = 0 \). This specific equation depends not only on the function \( y \) itself but also on its first and second derivatives. Third-order differential equations, like the one here, typically require information about the value of the function and its first two derivatives at a specific point to fully describe a solution.

Understanding these equations helps us predict and understand complex behavior in systems governed by the underlying principles expressed in the equation.

Initial Conditions

Initial conditions refer to the specific values that the solution of a differential equation must satisfy at a particular point. These are crucial because they ensure that the solution is unique and tailored to a particular problem.

In our exercise, we are given specific initial conditions for three solution functions, \( y_1 \), \( y_2 \), and \( y_3 \). For instance, \( y_1(1) = 0 \), \( y_1^{\prime}(1) = 1 \), and \( y_1^{\prime \prime}(1) = 1 \). These initial conditions specify the behavior of the function and its derivatives at \( t = 1 \).

By providing these conditions, we allow the differential equation to have one specific solution out of potentially infinite possibilities, thereby specifying the exact behavior of the system described by the differential equation.

Linear Independence

Linear independence is a key concept in determining whether a set of solutions to a differential equation forms a fundamental set. In simple terms, functions are linearly independent if no function in the set can be written as a combination of the others.

For the given problem, the solutions \( y_1 \), \( y_2 \), and \( y_3 \) must be linearly independent to form a fundamental set. This means that no solution can be derived from the others via any scalar multiple or sum. Linear independence ensures that the solution set spans the entire solution space of the differential equation.

This property is critical because if the solutions were dependent, they would not fully capture all possible behaviors described by the equation.

Wronskian Determinant

The Wronskian determinant is a mathematical tool used to check the linear independence of a set of functions. It involves calculating a determinant from the functions and their derivatives.

In our example, the Wronskian for the solutions \( y_1 \), \( y_2 \), and \( y_3 \) is given by:\[W(y_1, y_2, y_3) = \begin{vmatrix} y_1 & y_2 & y_3 \ y_1^{\prime} & y_2^{\prime} & y_3^{\prime} \ y_1^{\prime\prime} & y_2^{\prime\prime} & y_3^{\prime\prime} \end{vmatrix}\]

By substituting the initial conditions into the Wronskian and evaluating it at \( t = 1 \), we found that \( W(y_1, y_2, y_3)|_{t=1} = -1 \). Since the Wronskian is non-zero, the solutions are linearly independent. Thus, they form a fundamental set.

This means that they provide a complete solution to the differential equation, capturing all behaviors that the equation describes.