Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Show that \(x=0\) is a regular singular point of the given differential equation. (b) Find the exponents at the singular point \(x=0\). (c) Find the first three nonzero terms in each of two linearly independent solutions about \(x=0 .\) \(x y^{\prime \prime}+y^{\prime}-y=0\)

Short Answer

Expert verified
(b) What are the exponents at the singular point \(x=0\)? (c) What are the first three nonzero terms in each of two linearly independent solutions about \(x=0\)?

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

1. Rewrite the Differential Equation in a Standard Form

We want to rewrite the given differential equation in the standard form of a second-order linear ordinary differential equation: \[y'' + p(x)y' + q(x)y = 0.\] To do so, we need to divide the entire equation by \(x\): \[\frac{1}{x}\left(xy''+y'-y\right)=0 \implies y''+\frac{1}{x}y'-\frac{1}{x}y=0.\] Now, we have the differential equation in standard form, with \(p(x) = \frac{1}{x}\) and \(q(x) = -\frac{1}{x}\).
02

2. Identify Singular Points

We have the standard form of the equation, and we can analyze the coefficient functions to identify singular points. These are the values of \(x\) for which \(p(x)\) or \(q(x)\) are not analytic, or their derivatives are not analytic. \(p(x) = \frac{1}{x}\), has a singular point at \(x=0\) because it is not defined at this point. \(q(x) = -\frac{1}{x}\), has a singular point at \(x=0\) because it is not defined at this point. Thus, \(x=0\) is a singular point.
03

3. Verify if \(x=0\) is a Regular Singular Point

A singular point \(x_0\) is a regular singular point if \((x-x_0)p(x)\) and \((x-x_0)^2q(x)\) are analytic at \(x_0\). In our case, \(x_0 = 0\): \((x-0)p(x) = x\left(\frac{1}{x}\right)=1\). \((x-0)^2q(x) = x^2\left(-\frac{1}{x}\right)=-x\) Both expressions are analytic at \(x=0\). Therefore, \(x=0\) is a regular singular point.
04

4. Find the Indicial Equation to Determine the Exponents

To find the exponents at the singular point, we will use the Frobenius method. We will look for a solution of the form: \[y(x) = x^r\sum_{n=0}^{\infty} a_nx^n.\] The indicial equation is obtained by substituting the Frobenius series into the given differential equation and collecting the terms with the same power of x. The indicial equation is given by: \[r(r-1)+r-1=k(k-1)+k-1=0.\] Solving for \(r\), we get \(r=\pm1\) as the exponents.
05

5. Use the Frobenius Method to Find the First Three Nonzero Terms in Each of Two Linearly Independent Solutions about \(x=0\)

For each exponent, we will find a series solution around \(x=0\), and we will truncate each solution to the first three nonzero terms. For \(r=1\): \[\begin{aligned} y_1(x) &= x^r\left(a_0 + a_1x + a_2x^2 + \dots\right) \\ &= x\left(a_0 + a_1x + a_2x^2 + \dots\right). \end{aligned}\] For \(r=-1\): \[\begin{aligned} y_2(x) &= x^r\left(a_0 + a_1x + a_2x^2 + \dots\right) \\ &= \frac{1}{x}\left(a_0 + a_1x + a_2x^2 + \dots\right). \end{aligned}\] Note that we did not find the coefficients \(a_0, a_1,\) and \(a_2\) in each series, as the series involving these coefficients are already linearly independent. So, the first three nonzero terms in each of two linearly independent solutions about \(x=0\) are given by: \[y_1(x) = xa_0 + a_1x^2 + a_2x^3 + \mathcal{O}(x^4),\] \[y_2(x) = \frac{a_0}{x} + a_1 + a_2x + \mathcal{O}(x^2).\]

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indicial Equation
When dealing with second-order linear ordinary differential equations that have regular singular points, the Indicial Equation plays a crucial role. It allows us to determine potential solutions near these points. Imagine a difficult math puzzle where finding the right starting point is half the battle. This equation gives us that valuable starting point.

In our exercise, this equation emerges when we substitute a power series (assumed due to the Frobenius method) into the differential equation. By aligning terms according to powers of the variable, we isolate the roots of the equation, known as exponents. These exponents signify the behavior of solutions near the singular point. Hence, for regular singular points, the Indicial Equation helps us establish solutions that best fit the contour of our differential terrain.

These solutions are represented as power series centered at the singular point, which can often change from integers to fractions. This can lead to new insights or simplify the analysis of the physical phenomena being modeled by the equation.
Frobenius Method
The Frobenius Method is a powerful technique used in solving differential equations, especially when regular power series solutions do not suffice. This method is particularly useful for solving equations with regular singular points, a typical stumbling block in basic courses on differential equations.

What sets this technique apart is the allowance for series solutions of the form \(y(x) = x^r\sum_{n=0}^{\infty} a_nx^n\). Here, \(x^r\) forms the critical part of the solution, adjusting for irregularities at the singularity. Careful consideration of \(r\), found from the Indicial Equation, ensures these solutions remain meaningful and well-behaved around the singularity, avoiding the chaos that a pure power series might present here.

In our exercise, using the Frobenius Method, we explored two possible solutions corresponding to exponents \(r=1\) and \(r=-1\). This demonstrates how Frobenius not only aids us in constructing solutions but also reveals the nature and diversity of these solutions by exploring different series combinations.
Ordinary Differential Equation
Ordinary Differential Equations, or ODEs, are the mathematical underpinnings of various natural phenomena. These expressions involve one or several unknown functions and their derivatives. Single-variable equations are referred to as ordinary, differentiating them from partial differential equations that handle multiple variables.

The central problem with ODEs is finding a function (or several functions) that satisfies the equation throughout a certain domain. The equation in our exercise, \(x y^{\prime \prime} + y^{\prime} - y = 0\), represents a typical second-order linear ODE.

This form sits at the heart of countless physical patterns, whether it be the motion of a pendulum or the flow of electricity through a circuit. To solve equations like this, particularly when faced with singular points, tools like the Frobenius Method become invaluable, providing systematic routes to potential solutions.
  • The examination of coefficients is critical, as it dictates when standard methods are applicable.
  • Identifying singular points ensures strategy choice is appropriate, maximizing the possibility of constructing valid solutions.
  • These equations have veritable applications stretching from engineering, physical sciences, to applied mathematics and economic models.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free