Question

verify that the functions \(y_{1}\) and \(y_{2}\) are solutions of the given differential equation. Do they constitute a fundamental set of solutions? $$ (1-x \cot x) y^{\prime \prime}-x y^{\prime}+y=0, \quad 0

Short answer

Based on the provided step-by-step solution, we can create the following short answer: Both \(y_1(x) = x\) and \(y_2(x) = \sin x\) are solutions to the given differential equation \((1-x \cot x) y^{\prime\prime}-x y^{\prime}+y=0\). The Wronskian, \(W(y_1, y_2) = x\cos x - \sin x\), is not identically zero in the domain \(0 < x < \pi\). Therefore, \(y_1(x) = x\) and \(y_2(x) = \sin x\) form a fundamental set of solutions to the given differential equation.

Step-by-step solution

Analyze the given differential equation

The given differential equation is \((1-x \cot x) y^{\prime\prime}-x y^{\prime}+y=0\). Our task is to verify if \(y_1(x) = x\) and \(y_2(x) = \sin x\) are solutions to this differential equation.

Substitute \(y_1\) and check if it satisfies the differential equation

Let's take \(y_1(x) = x\) and calculate its first and second derivatives: \(y_1^{\prime}(x) = 1\), \(y_1^{\prime\prime}(x)= 0\). Now, let's substitute \(y_1(x)\), \(y_1^{\prime}(x)\), and \(y_1^{\prime\prime}(x)\) into the differential equation: \((1-x \cot x)(0)-x(1)+x=0\). Simplify the expression: \(x - x = 0\). This is a true statement, so \(y_1(x) = x\) is a solution to the given differential equation.

Substitute \(y_2\) and check if it satisfies the differential equation

Let's take \(y_2(x) = \sin x\) and calculate its first and second derivatives: \(y_2^{\prime}(x) = \cos x\), \(y_2^{\prime\prime}(x)= -\sin x\). Now, let's substitute \(y_2(x)\), \(y_2^{\prime}(x)\), and \(y_2^{\prime\prime}(x)\) into the differential equation: \((1-x \cot x)(-\sin x)-x(\cos x)+\sin x=0\). Multiply and simplify the expression: \(-\sin x + x\sin x + x\cos x + \sin x = 0\). This is a true statement, so \(y_2(x) = \sin x\) is also a solution to the given differential equation.

Determine if \(y_1\) and \(y_2\) form a fundamental set of solutions

To check if \(y_1(x)\) and \(y_2(x)\) form a fundamental set of solutions, we need to compute their Wronskian: \(W(y_1, y_2) = \begin{vmatrix}y_1 & y_2 \\ y_1^{\prime} & y_2^{\prime}\end{vmatrix} = \begin{vmatrix}x & \sin x \\ 1 & \cos x \end{vmatrix}\). Now, calculate the determinant: \(W(y_1, y_2) = x\cos x - \sin x\). The function \(W(y_1, y_2) = x\cos x - \sin x\) is not identically zero in the domain \(0 < x < \pi\). Thus, the given functions \(y_1(x) = x\) and \(y_2(x) = \sin x\) form a fundamental set of solutions to the given differential equation.

Key concepts

Fundamental Set of Solutions

Understanding the fundamental set of solutions is critical when dealing with homogeneous linear differential equations. It's about finding a group of solutions that can represent all possible solutions. With these 'building blocks,' any other solution of the differential equation can be expressed as a linear combination of the elements in this set.

In the exercise, we analyzed if the functions \(y_1(x) = x\) and \(y_2(x) = \sin x\) are part of such a set for the differential equation \( (1-x \cot x) y^{\prime \prime}-x y^{\prime}+y=0\). After verifying that both functions satisfy the differential equation independently, it was essential to determine if they are linearly independent. This is done using the Wronskian, which we will discuss next. If the Wronskian of these functions is not zero, it indicates that they are indeed linearly independent and, therefore, constitute a fundamental set of solutions. This fundamental set provides us with a powerful tool to describe all solutions to the differential equation in the given domain \(0

Wronskian

The Wronskian is a concept named after the Polish mathematician Józef Hoene-Wronski. It is a method used to determine whether a set of functions is linearly independent, an essential aspect of finding the fundamental set of solutions. By calculating the Wronskian of two functions, we are essentially taking the determinant of a matrix composed of these functions and their derivatives.

In our exercise, we calculated the Wronskian of \(y_1\) and \(y_2\) as follows:
\begin{align*} W(y_1, y_2) &= \begin{vmatrix} y_1 & y_2 \ y_1^{\prime} & y_2^{\prime}\end{vmatrix} = \begin{vmatrix} x & \sin x \ 1 & \cos x \end{vmatrix} \ &= x\cos x - \sin x.\end{align*}
The result, \(W(y_1, y_2) = x\cos x - \sin x\), is not constantly zero in the interval \(0

Solving Differential Equations

Solving differential equations is a staple of mathematical study and application, with techniques varying widely based on the type of equation at hand. Whether dealing with ordinary differential equations (ODEs) or partial differential equations (PDEs), the solutions can describe a multitude of physical phenomena, such as sound waves, heat conduction, or population growth.

In the textbook exercise, we focused on solving a specific type of ODE, verifying given solutions, and then determining their independence using the Wronskian. The successful verification process not only confirmed our solutions but also empowered us to use the fundamental set of solutions for constructing general solutions for similar differential equations within the given domain. The ability to solve such equations and understand their solutions' structure allows us to model and predict a vast range of real-world situations—underlining the power and necessity of mastery in solving differential equations.

Wronskian

The Wronskian is a concept named after the Polish mathematician Józef Hoene-Wronski. It is a method used to determine whether a set of functions is linearly independent, an essential aspect of finding the fundamental set of solutions. By calculating the Wronskian of two functions, we are essentially taking the determinant of a matrix composed of these functions and their derivatives.

In our exercise, we calculated the Wronskian of \(y_1\) and \(y_2\) as follows:
\begin{align*} W(y_1, y_2) &= \begin{vmatrix} y_1 & y_2 \ y_1^{\prime} & y_2^{\prime}\end{vmatrix} = \begin{vmatrix} x & \sin x \ 1 & \cos x \end{vmatrix} \ &= x\cos x - \sin x.\end{align*}
The result, \(W(y_1, y_2) = x\cos x - \sin x\), is not constantly zero in the interval \(0

Solving Differential Equations

Solving differential equations is a staple of mathematical study and application, with techniques varying widely based on the type of equation at hand. Whether dealing with ordinary differential equations (ODEs) or partial differential equations (PDEs), the solutions can describe a multitude of physical phenomena, such as sound waves, heat conduction, or population growth.

In the textbook exercise, we focused on solving a specific type of ODE, verifying given solutions, and then determining their independence using the Wronskian. The successful verification process not only confirmed our solutions but also empowered us to use the fundamental set of solutions for constructing general solutions for similar differential equations within the given domain. The ability to solve such equations and understand their solutions' structure allows us to model and predict a vast range of real-world situations—underlining the power and necessity of mastery in solving differential equations.

Solving Differential Equations

Solving differential equations is a staple of mathematical study and application, with techniques varying widely based on the type of equation at hand. Whether dealing with ordinary differential equations (ODEs) or partial differential equations (PDEs), the solutions can describe a multitude of physical phenomena, such as sound waves, heat conduction, or population growth.

In the textbook exercise, we focused on solving a specific type of ODE, verifying given solutions, and then determining their independence using the Wronskian. The successful verification process not only confirmed our solutions but also empowered us to use the fundamental set of solutions for constructing general solutions for similar differential equations within the given domain. The ability to solve such equations and understand their solutions' structure allows us to model and predict a vast range of real-world situations—underlining the power and necessity of mastery in solving differential equations.