# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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lim (x - x0) Q^ ) is finite (6)

X^X0 P (x )

and

lim (x - x0)2—— is finite. (7)

)2 R(x ) X0) P (x)

This means that the singularity in Q/P can be no worse than (x - x0)—1 and the singularity in R/P can be no worse than (x - x0)—2. Such a point is called a regular singular point of Eq. (1). For more general functions than polynomials, x0 is a regular singular point of Eq. (1) if it is a singular point, and if both9

( ) Q(X) d ( \2 R(X)

(x - x0) and (x - x0) (8)

0

9The functions given in Eq. (8) may not be defined at xq, in which case their values at xq are to be assigned as their limits as x ^ xq.

258

Chapter 5. Series Solutions of Second Order Linear Equations

have convergent Taylor series about x0, that is, if the functions in Eq. (8) are analytic at x = x0. Equations (6) and (7) imply that this will be the case when P, Q, and R are polynomials. Any singular point of Eq. (1) that is not a regular singular point is called an irregular singular point of Eq. (1).

In the following sections we discuss how to solve Eq. (1) in the neighborhood of a regular singular point. A discussion of the solutions of differential equations in the neighborhood of irregular singular points is more complicated and may be found in more advanced books.

In Example 2 we observed that the singular points of the Legendre equation

(1 — x2) y" — 2xy' + a(a + 1)y = 0

are x = ±1. Determine whether these singular points are regular or irregular singular points.

We consider the point x = 1 first and also observe that on dividing by 1 — x2 the coefficients of y' and y are —2x/(1 — x2) and a(a + 1)/(1 — x2), respectively. Thus we calculate

—2x (x — 1)(—2x) 2x

lim (x — 1)---------~ = lim-------------------= lim--------= 1

x ^1 1 — x x ^1 (1 — x )(1 + x) x^1 1 + x

and

2 a(a + 1) (x — 1)2a(a + 1)

lim (x — 1) --------= lim---------------------------

x ^1 1 — x x^1 (1 — x )(1 + x)

(x — 1)(—a)(a + 1) A

= lim--------------------------= 0.

x^ 1 1 + x

Since these limits are finite, the point x = 1 is a regular singular point. It can be shown in a similar manner that x = — 1 is also a regular singular point.

EXAMPLE

6

Determine the singular points of the differential equation

2x (x — 2)2 y" + 3xy; + (x — 2) y = 0 and classify them as regular or irregular.

Dividing the differential equation by 2x (x — 2)2, we have

31

y +----------î y +---------y = 0,

2(x — 2)2 2x (x — 2)

so p(x) = Q(x)/P(x) = 3/2(x — 2)2 and q(x) = R(x)/P(x) = 1/2x(x — 2). The

singular points are x = 0 and x = 2. Consider x = 0. We have

3

lim xp(x) = lim x-------~ = 0,

x^0 ^ x^0 2(x — 2)2

lim x2q (x) = lim x2-------= 0.

2x (x — 2)

5.4 Regular Singular Points

259

EXAMPLE

Since these limits are finite, x — 0 is a regular singular point. For x — 2 we have

3 . 3

lim (x - 2)p(x) — lim (x - 2) —---------— — lim

x^2 x^2 2(x - 2)2 x^2 2(x - 2)

so the limit does not exist; hence x — 2 is an irregular singular point.

Determine the singular points of

/ Ï \ 2

7 (x - 2) y" + (cos x)y'+ (sin x)y — 0

and classify them as regular or irregular.

The only singular point is x — ï/2. To study it we consider the functions

ï \ / ï \ Q (x) cos x

(x - ’-) p(x> — (x - 2)

2' P (x) x - ï/2

and

ï\2 / ï\2 R(x)

sin x.

2/ V 2) P (x)

Starting from the Taylor series for cos x about x — ï/2, we find that

cos x (x - ï/2)2 (x - ï/2)4

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