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

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ri = 0, r2 = -1. Treating an as a function of r, we see

that an(r) = -an-1(r)/F(r+n), n = 1,2,... if F(r+n) Ô 0.

Thus a1(r) = -a0/F(r+1), a2(r) = a0/F(r+1)F(r+2),..., and

an(r) = (-1) na0/F(r+1)F(r+2)...F(r+n), provided F(r+n) Ô

0 for n = 1,2,... . For the case r1 = 0, we have

an(0) = (-1)na0/F(1)F(2) ... F(n) = (-1)na0/n!(n+1)! so

one solution is y1(x) = (-1) nxn/n!(n+1)! where we have

n=0

set a0 = 1.

If we try to use the above recurrence relation for

Section 5.8

99

the case r2 = -1 we find that an(-1) = -an_1/n(n-1), which is undefined for n = 1. Thus we must follow the procedure described at the end of Section 5.7 to

calculate a second solution of the form given in Eq.(24).

Specifically, we use Eqs.(19) and (20) of that section to calculate a and cn(r2), where r2 = -1. Since

r1 - r2 = 1 = N, we have aN(r) = a1(r) = -1/F(r+1), with

a0 = 1. Hence

a = lim [(r+1)(-1)/F(r+1)] = lim [-(r+1)/(r+1)(r+2)] = -1.

r ^ -1 r ^ -1

Next

d

cn(-1) = —[(r+1)an(r)] dr

, (-1) -----------------<?+!>--------

_ dr F(r+1) ... F(r+n)

r=-1

r=-1

where we again have set a0 = 1. Observe that

2 2 2

(r+1)/F(r+1)... F(r+n)=1/[(r+2) (r+3) ...(r+n) (r+n+1)]=1/Gn(r) Hence cn(-1) = (-1) n+1Gn(-1)/Gn(-1). Notice that Gn(-1) = 12-22-32 ...(n-1)2n = (n-1)!n! and

Gn(-1)/Gn(-1) = 2[1/1 + 1/2 + 1/3 +...+ 1/(n-1)] + 1/n =

Hn + Hn-1. Thus cn(-1) = (-1) n+1(Hn + Hn-1)/(n-1)!n!.

From Eq.(24) of Section 5.7 we obtain the second solution

to

Ó2 (x) = - y1(x)lnx + x-1[1 - X (-1)n(Hn + Hn-1)xn/n!(n-1)!].

n=1

2. 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

xp(x) = 3 and x q(x) = 1+x. Both are analytic at x = 0,

so x = 0 is a regular singular point. Substituting

to

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

n=0

appropriately, and collecting coefficients of like powers of x yields

to

[r(r-1) + 3r + 1]a0xr + {[(r+n)(r+n+2) + 1]an

n=1

n+r

+ an-1} x = 0.

22 The indicial equation is F(r) = r + 2r + 1 = (r+1) = 0

with the double root r1 = r2 = -1. Treating an as a function of r, we see that an(r) = -an-1(r)/F(r+n),

100

Section 5.8

n = 1,2,... . Thus a1(r) = -a0/F(r+1),

a2(r) = a0/F(r+1)F(r+2),..., and

an(r) = (-1) na0/F(r+1)F(r+2)... F(r+n). Setting r = -1 we find that an(-1) = (-1) na0/(n!)2, n = 1,2,... . Hence one

ro

solution is y1(x) = x-1I (-1)nxn/(n!)2 where we have set

n=0

a0 = 1. To find a second solution we follow the

procedure described in Section 5.7 for the case when the roots of the indicial equation are equal. Specifically, the second solution will have the form given in Eq.(17)

of that section. We must calculate an(-1). If we let

2 2 2

Gn(r) = F(r+1)...F(r+n) = (r+2) (r+3) ... (r+n+1) and

take a0 = 1, then a^-1) = (-1) n[1/Gn(r)]' evaluated

r = -1. Hence a'n(-1) = (-1)n+1G^ (-1)/Gn (-1). But

Gn(-1) = (n!)2 and Gn(-1)/Gn(-1) = 2[1/1 + 1/2 + 1/3 + ... + 1/n] = 2Hn. Thus a second solution is

ro

y2 (x) = y1 (x)lnx - 2x-1I (-1)nHnxn/(n!)2.

n=1

3. The roots of the indicial equation are r1 and r2 = 0 and

thus the analysis is similar to that for Problem 2.

4. The roots of the indicial equation are r1 = -1 and

r2 = -2 and thus the analysis is similar to that for

Problem 1.

5. Since x = 0 is a regular singular point, substitute

ro

y = I anxn+r in the D.E., shift indices appropriately,

n=0

and collect coefficients of like powers of x to obtain [r2 - 9/4]a0xr + [(r+1)2 - 9/4]a1xr+1 ro

+ I {[(r+n)2 - 9/4]an + an-2} xn+r = 0.

n=2

2

The indicial equation is F(r) = r - 9/4 = 0 with roots r1 = 3/2, r2 = -3/2. Treating an as a function of r we see that an(r)= -an-2(r)/F(r+n), n = 2,3,.. if F(r+n) Ô

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