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Chapter 8: Problem 6

In Problems, I through 8, determine whether \(x=0\) is an ordinary point, a regular singular point, or an irregular singular point. If it is a regular singular point, find the exponents of the differential equation at \(x=0 .\) $$ x^{2}\left(1-x^{2}\right) y^{\prime \prime}+2 x y^{\prime}-2 y=0 $$

Short Answer

Expert verified
The point \( x = 0 \) is a regular singular point with exponents \( r = 0 \) and \( r = -1 \).

Step by step solution

01

Identify Coefficients

The standard form for a second order linear differential equation is \( P(x)y'' + Q(x)y' + R(x)y = 0 \). For this equation, \( P(x) = x^2(1-x^2) \), \( Q(x) = 2x \), and \( R(x) = -2 \).
02

Analyze the Singular Points

A singular point occurs if \( P(x) = 0 \). For our equation, the singular points occur at \( x = 0 \) and \( x = \pm 1 \) since \( P(x) = x^2(1-x^2) \). However, here we will focus on \( x = 0 \).
03

Determine the Type of Singularity at \( x = 0 \)

To determine the type, compare the limits of \( \frac{Q(x)}{P(x)} \) and \( \frac{R(x)}{P(x)} \) as \( x \to 0 \).\[ \frac{Q(x)}{P(x)} = \frac{2x}{x^2(1-x^2)} = \frac{2}{x(1-x^2)} \] \[ \frac{R(x)}{P(x)} = \frac{-2}{x^2(1-x^2)} \]Both expressions have leading terms with negative power of \( x \), which indicates the point is singular.
04

Confirm Regular Singular Point

For a regular singular point at \( x = 0 \), \( x\frac{Q(x)}{P(x)} \) and \( x^2\frac{R(x)}{P(x)} \) should both have finite limits as \( x \to 0 \).Compute:\[ x \frac{Q(x)}{P(x)} = \frac{2}{(1-x^2)} \to 2 \text{ (finite)} \]\[ x^2 \frac{R(x)}{P(x)} = \frac{-2x}{1-x^2} \to 0 \text{ (finite)} \]Both limits are finite indicating \( x = 0 \) is a regular singular point.
05

Find Exponents Using Indicial Equation

The indicial equation for a generic regular singular point \( x = 0 \) is:\[ r(r-1) + a_0 r + b_0 = 0 \]Here, \( a_0 = 2 \) (from \( x \frac{Q(x)}{P(x)} \)) and \( b_0 = 0 \) (from \( x^2 \frac{R(x)}{P(x)} \)).The equation becomes:\[ r^2 + r = 0 \]Solve for \( r \):\[ r(r+1) = 0 \]The exponents are \( r = 0 \) and \( r = -1 \).

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Key Concepts

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

Ordinary Point
An ordinary point in the context of differential equations is a point where the coefficients do not lead to any form of singularity. In a second order linear differential equation, this means that if we have an equation of the form \( P(x)y'' + Q(x)y' + R(x)y = 0 \), the point \( x=a \) would be an ordinary point if the function \( P(x) \) is not zero at \( x=a \). Ordinary points are crucial because they allow solutions to be expressed in terms of a power series around that point, facilitating easy calculation of differential equation solutions. If a differential equation has only ordinary points, it simplifies the path towards finding explicit solutions.
Regular Singular Point
A regular singular point occurs when \( P(x) = 0 \) but the limits \( x \frac{Q(x)}{P(x)} \) and \( x^2 \frac{R(x)}{P(x)} \) are finite as \( x \to a \). Regular singular points involve some level of complexity, but solutions can still potentially take a power series form, often requiring special functions known as Frobenius methods for solving. In the given exercise, \( x = 0 \) is distinguished as a regular singular point. By simplifying the coefficients' limits, it was verified that this point meets the finite limit condition for both expressions, confirming its regular singular nature.
Indicial Equation
The indicial equation is essential when dealing with regular singular points, as it helps determine the allowable forms of the solution near these points. Essentially, it arises from the method of Frobenius, which is used to find series solutions to differential equations at singular points. The indicial equation is typically derived from substituting a trial solution into the differential equation and equating the lowest power of \( x \) terms to zero. In the provided solution, this equation took the form \( r^2 + r = 0 \), where the solutions \( r = 0 \) and \( r = -1 \) represented the exponents in the series solution.
Second Order Linear Differential Equation
Second order linear differential equations are characterized by the presence of the highest derivative being the second derivative. The general form is \( P(x)y'' + Q(x)y' + R(x)y = 0 \), where \( P(x), Q(x), \) and \( R(x) \) are functions of \( x \). These equations are pivotal in modeling various phenomena, including mechanical vibrations, electrical currents in circuits, and natural growth processes. Identifying points of singularity in these equations is crucial in determining how to approach solving them, whether through power series methods or other approaches when dealing with ordinary or singular points.

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Most popular questions from this chapter

Either apply the method of Example 1 to find two linearly independent Frobenius series solutions, or find one such solution and show (as in Example 2) that a second such solution does not exist. \(x y^{\prime \prime}+(5+3 x) y^{\prime}+3 y=0\)

Express the general solution of the given differential equation in terms of Bessel functions. $$ x y^{\prime \prime}-y^{\prime}+36 x^{3} y=0 $$

If \(x=a \neq 0\) is a singular point of a second-order linear differential equation, then the substitution \(t=x-a\) transforms it into a differential equation having \(t=0\) as a singular point. We then attribute to the original equation at \(x=a\) the behavior of the new equation at \(t=0 .\) Classify ( as regular or irregular) the singular points of the differential equations in Problems 9 through 16. $$ (x-2)^{2} y^{\prime \prime}-\left(x^{2}-4\right) y^{\prime}+(x+2) y=0 $$

Consider the equation \(x^{2} y^{\prime \prime}+x y^{\prime}+(1-x) y=0 .\) (a) Show that its exponents are \(\pm i\), so it has complex-valued Frobenius series solutions $$ y_{+}=x^{i} \sum_{n=0}^{\infty} p_{n} x^{n} \quad \text { and } \quad y_{-}=x^{-i} \sum_{n=0}^{\infty} q_{n} x^{n} $$ with \(p_{0}=q_{0}=1\). (b) Show that the recursion formula is $$ c_{n}=\frac{c_{n-1}}{n^{2}+2 r n} . $$ Apply this formula with \(r=i\) to obtain \(p_{n}=c_{n}\), then with \(r=-i\) to obtain \(q_{n}=c_{n}\). Conclude that \(p_{n}\) and \(q_{n}\) are complex conjugates: \(p_{n}=a_{n}+i b_{n}\) and \(q_{n}=a_{n}-i b_{n}\), where the numbers \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) are real. (c) Deduce from part (b) that the differential equation given in this problem has real- valued solutions of the form $$ \begin{aligned} &y_{1}(x)=A(x) \cos (\ln x)-B(x) \sin (\ln x), \\ &y_{2}(x)=A(x) \sin (\ln x)+B(x) \cos (\ln x) \end{aligned} $$ where \(A(x)=\sum a_{n} x^{n}\) and \(B(x)=\sum b_{n} x^{n}\).

The Hermite equation of order \(\alpha\) is \(y^{\prime \prime}-2 x y^{\prime}+2 \alpha y=0\). (a) Derive the two power series solutions $$ y_{1}=1+\sum_{m=1}^{\infty}(-1)^{m} \frac{2^{m} \alpha(\alpha-2) \cdots(\alpha-2 m+2)}{(2 m) !} x^{2 m} $$ and \(y_{2}=x\) $$ +\sum_{m=1}^{\infty}(-1)^{m} \frac{2^{m}(\alpha-1)(\alpha-3) \cdots(\alpha-2 m+1)}{(2 m+1) !} x^{2 m+1}. $$ Show that \(y_{1}\) is a polynomial if \(\alpha\) is an even integer, whereas \(y_{2}\) is a polynomial if \(\alpha\) is an odd integer. (b) The Hermite polynomial of degree \(n\) is denoted by \(H_{n}(x) .\) It is the \(n\) th-degree polynomial solution of Hermite's equation, multiplied by a suitable constant so that the coefficient of \(x^{n}\) is \(2^{n}\). Show that the first six Hermite polynomials are $$ \begin{array}{llrl} H_{0}(x) & \equiv 1, & & H_{1}(x)=2 x, \\ H_{2}(x) & =4 x^{2}-2, & & H_{3}(x)=8 x^{3}-12 x, \\ H_{4}(x) & =16 x^{4}-48 x^{2}+12, & & \\ H_{5}(x) & =32 x^{5}-160 x^{3}+120 x & & \end{array} $$ A general formula for the Hermite polynomials is $$ H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right). $$ Verify that this formula does in fact give an \(n\) th-degree polynomial. It is interesting to use a computer algebra system to investigate the conjecture that (for each \(n)\) the zeros of the Hermite polynomials \(H_{n}\) and \(H_{n+1}\) are "interlaced" - that is, the \(n\) zeros of \(H_{n}\) lie in the \(n\) bounded open intervals whose endpoints are successive pairs of zeros of \(H_{n+1}\).
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