# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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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

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

?

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

n=0

gives

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

n+r+1-s

(n+r)arp-

n=0 n=0

98

Section 5.8

P X ' n+r+2-t ~

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

jr ? I -i \ r r+1-s , n 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

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) ^ 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) ^

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

r2 - r2 = 1 = N, we have aN(r) = a2(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) ğA,---------------<?h>----------

_ 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) ^^ğ(-D/G^-D. Notice that Gn(-1) = 12-22-32 ...(n-1)2n = (n-1)!n! and

Gğ(-1)/Gğ(-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

y2 (x) = - yx(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

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

n=0

appropriately, and collecting coefficients of like powers of x yields

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

n=1

n+r

+ ^-1/ x = °.

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 = 1,9,... . Thus a1(r) = -a0/F(r+1),

a9(r) = a0/F(r+1)F(r+9),..., and

an(r) = (-1) na0/F(r+1)F(r+9)... F(r+n). Setting r = -1 we find that an(-1) = (-1) na0/(n!)9, n = 1,9,... . Hence one

solution is y1(x) = x-1^ (-1)nxn/(n!)9 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

9 9 9

Gn(r) = F(r+1)...F(r+n) = (r+9) (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+1Gn(-1)/Gn;(-1). But

Gn(-1) = (n!)9 and Gn(-1)/Gn(-1) = 9[1/1 + 1/9 + 1/3 + ... + 1/n] = 9Hn. Thus a second solution is

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