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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
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.
2 0b. Substituting y = xr1 anxn in x3y" + axy' + Py = 0 gives
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
+ (n+r-1)(n+r-2)an-1 + [a(n+r) + p]anxn+r = 0. Thus
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.
For — = -1, then, an = -----------------, which is zero for
a n an
n = 1 and thus y(x) = x is the solution. Similarly for
— = 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
ro as n ^ ro and thus the ratio test yields a zero radius of convergence.
21b. Substituting y = anxn+r in the D.E. in standard form
11 (n+r)(n+r-1)anxn+r + a (n+^a^
PV n+r+2-t ë
X anX = 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
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
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.
The indicial equation is F(r) = r(r+1) = 0 with roots