Question

Find the Wronskian of a given set \(\left\\{y_{1}, y_{2}\right\\}\) of solutions of $$y^{\prime \prime}+3\left(x^{2}+1\right) y^{\prime}-2 y=0$$ given that \(W(\pi)=0\).

Short answer

Answer: The evaluated Wronskian at \(x = \pi\) is \(W(\pi)= y_1(\pi)y_2'(\pi) - y_1'(\pi)y_2(\pi) = 0\).

Step-by-step solution

Form the Wronskian determinant

Since we want to find the Wronskian of the two given solutions, we form the determinant of their derivatives: $$ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} $$

Use the given differential equation to solve for \(y''\)

We have: $$ y^{\prime \prime} + 3\left(x^{2}+1\right) y^{\prime} -2y=0 $$ Solve for \(y''\): $$ y^{\prime \prime} = -3(x^2 + 1)y' + 2y $$

Determine the derivatives \(y_1', y_2', y_1'', y_2''\) at \(x = \pi\)

We currently don't have information to find exact forms of \(y_1\) and \(y_2\) or their derivatives directly. However, we can represent the derivatives of \(y_1\) and \(y_2\) at \(x = \pi\) in terms of \(y_1(\pi)\), \(y_2(\pi)\), \(y_1'(\pi)\), and \(y_2'(\pi)\) using the differential equation. Thus, at \(x= \pi\), we have the following derivatives: $$ y_1''(\pi) = -3(\pi^2 + 1)y_1'( \pi ) + 2y_1( \pi ) $$ $$ y_2''(\pi) = -3(\pi^2 + 1)y_2'( \pi ) + 2y_2( \pi ) $$

Evaluate the Wronskian determinant at \(x = \pi\) with given information

Given that the Wronskian is \(0\) at \(x=\pi\), we form the determinant at \(x=\pi\): $$ W(\pi) = \begin{vmatrix} y_1(\pi) & y_2(\pi) \\ y_1'(\pi) & y_2'(\pi) \end{vmatrix} = 0 $$ Expanding the determinant, we get: $$ W(\pi)= y_1(\pi)y_2'(\pi) - y_1'(\pi)y_2(\pi) = 0 $$

Conclusion

We've found the Wronskian of the given set of solutions at \(x = \pi\): $$ W(\pi)= y_1(\pi)y_2'(\pi) - y_1'(\pi)y_2(\pi) = 0 $$ Although we couldn't find the exact forms of the given functions and their derivatives, we're able to relate them based on the given differential equation and show the relationship at \(x = \pi\).

Key concepts

Understanding Differential Equations

Differential equations are mathematical equations that describe how a quantity changes in relation to another. They are a cornerstone of modern science and engineering, providing a way to model everything from the motions of heavenly bodies to the fluctuations of financial markets. A particularly important class of differential equations is second-order linear differential equations, which can be written in the form \(y'' + p(x)y' + q(x)y = 0\). These equations come with two constants of integration since they are second-order, implying that they have two independent solutions. The Wronskian, associated with the given equation in the exercise, serves to determine if a set of solutions forms a fundamental set, which is necessary for the broader solution of the differential equation.

In the provided exercise, the differential equation \(y''+3(x^2+1)y'-2y=0\) is a homogeneous second-order differential equation, which means all the terms are proportional to either \(y\) or one of its derivatives, and it equals to zero. The goal is to find the Wronskian to ascertain the relationship between the solutions \(y_1\) and \(y_2\). The process involves calculating the derivatives and using a determinant to find the Wronskian, as shown in the steps of the solution.

The Role of Boundary Value Problems

Boundary Value Problems (BVPs) are a type of differential equation problem where the solution is required to satisfy certain conditions, called 'boundary conditions', at the endpoints of the interval on which the equation is defined. BVPs are distinct from initial value problems, which only require information about the solution at a single point (the initial condition).

BVPs are essential in physics and engineering disciplines, for instance, when determining the temperature distribution along a rod over time or the deflection of a beam subject to certain loads. In our situation, the Wronskian is evaluated at an endpoint of our interval: at \(x=\pi\). The boundary condition provided, \(\textbf{W}(\pi)=0\), is a crucial piece of information that helps us calculate the Wronskian at \(x = \pi\). It's important to realize that the Wronskian being zero at this point suggests a certain dependency between the solutions \(y_1\) and \(y_2\). Furthermore, in more complex scenarios, the boundary conditions might not be as straightforward as \(W(x)=0\), but could involve other relationships between the solutions and their derivatives.

Determinants: Key to the Wronskian

The determinant is a powerful mathematical tool that transforms a square matrix into a single number. This number encapsulates important properties about the linear independence of vectors or, in the context of differential equations, the independence of solutions. Determinants also appear in a variety of mathematical and physical contexts, such as computing volumes in higher dimensions and solving systems of linear equations.

For our task, the determinant is used to compute the Wronskian, which is the determinant of a matrix comprising the functions \(y_1\), \(y_2\) and their first derivatives \(y_1'\), \(y_2'\). The Wronskian itself is a function that can indicate whether the given set of solutions is linearly independent. If the Wronskian is non-zero at any point within the interval of interest, then the solutions are linearly independent. In the context of the exercise, we specifically looked at \(W(\pi)\), which was found to be zero and thus implies, that at \(x=\pi\), the solutions \(y_1\) and \(y_2\) may not be linearly independent. This illustrates an inherent relationship between the solutions that can heavily influence the structure of the problem's general solution.