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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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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) 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
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
2 (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)
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