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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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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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