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

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better to keep the differential equation in its original

form and to substitute the above power series for lnx:

1 2 1 3 1 4 1 ,

[(x-1) - - (x-1) 2 + - (x-1) 3 - - (x-1) 4 + ...]y" + -y' + y = 0.

2 3 4 2

Next we substitute y = a0(x-1) 1/2 + a1(x-1) 3/2 + a2(x-1) 5/2

+ ... and collect coefficients of like powers of (x-1) which are then set equal to zero. This requires some algebra before we find that 6a1/4 + 9a0/8 = 0 and

5a2 + 5a1/8 - a0/12 = 0. These equations yield

a1 = -3a0/4 and a2 = 53a0/480. With a0 = 1 we obtain the

solution

1/2 3 3/2 53 5/2

y1(x) = (x-1) - — (x-1) + -(x-1) + ... . Since

1 4 480

the radius of convergence of the series for ln x is 1, we

would expect p = 1.

20a. If we write the D.E. in the standard form as given in

Theorem 5.7.1 we obtain x2y" + x[a/x]y' + [0/x]y = 0 where

xp(x) = a/x and x2q(x) = â/x. Neither of these terms are

analytic at x = 0 so x = 0 is an irregular singular point.

Section 5.7

97

2 0b. Substituting y = xr1 anxn in x3y" + axy' + Py = 0 gives

n=0

ÃÎ ÃÎ ÃÎ

X ’ , x , .. x n+r+1 ë X ’ , x n+r n X’ n+r ~

I (n+r)(n+r-1)anx + a I (n+^a^ + p I a^ = 0.

n=0 n=0 n=0

Shifting the index in the first series and collecting coefficients of common powers of x we obtain (ar + P)a0xr

ro

+ (n+r-1)(n+r-2)an-1 + [a(n+r) + p]anxn+r = 0. Thus

n=1

the indicial equation is ar + p = 0 with the single root r = - p/a.

20c. From part b, the recurrence relation is (n+r-1)(n+r-2)an-1

an = --------------- -------, n = 1,2,...

a(n+r) + P pp (n -— -1)(n - — -2)an-1 aa

--------, for r = -p/a.

an

p n(n-1)an-1

For — = -1, then, an = -----------------, which is zero for

a n an

n = 1 and thus y(x) = x is the solution. Similarly for

p (n-1)(n-2)

— = 0, an = - and again for n = 1 a1 = 0 and

a n an 1

y(x) = 1 is the solution. Continuing in this fashion, we see that the series solution will terminate for p/a any

positive integer as well as 0 and -1. For other values

pp a (n- — -1)(n-— -2)

an 2 a

of p/a, we have -------- = --------------------, which approaches

an-1 an

ro as n ^ ro and thus the ratio test yields a zero radius of convergence.

ro

21b. Substituting y = anxn+r in the D.E. in standard form

n=0

gives

11 (n+r)(n+r-1)anxn+r + a (n+^a^

n+r+1-s

(n+rja^

n=0 n=0

98

Section 5.8

PV n+r+2-t ë

X anX = 0.

n=0

If s = 2 and t = 2 the first term in each of the three series is r(r-1)a0xr, ara0xr-1, and Pa0xr, respectively. Thus we must have ara0 = 0 which requires r = 0. Hence there is at most one solution of the assumed form.

21d. In order for the indicial equation to be quadratic in r

it is necessary that the first term in the first series contribute to the indicial equation. This means that the first term in the second and the third series cannot appear before the first term of the first series. The

JT ¦ ä_ ä_ I -I \ r r+1-s , Q r + 2-t

first terms are r(r-1)a0x , ara0x , and pa0x ,

respectively. Thus if s < 1 and t < 2 the quadratic term will appear in the indicial equation.

Section 5.8, Page 289

1. It is clear that x = 0 is a singular point. The D.E. is

in the standard form given in Theorem 5.7.1 with

2

xp(x) = 2 and x q(x) = x. Both are analytic at x = 0, so

x = 0 is a regular singular point. Substituting

y = anxn+r in the D.E., shifting indices

n=0

appropriately, and collecting coefficients of like powers of x yields

[r(r-1) + 2r]a0xr + [(r+n)(r+n+1)an + an-1]xr+n = 0.

n=1

The indicial equation is F(r) = r(r+1) = 0 with roots

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