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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.
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We cannot say more without further analysis.
Observe that no complicated calculations were required to discover the information about the solutions presented in this example. All that was needed was to evaluate a few limits and solve two quadratic equations.
We now consider the cases in which the roots of the indicial equation are equal, or differ by a positive integer, r1 — r2 = N. As we have shown earlier, there is always one solution of the form (9) corresponding to the larger root r1 of the indicial equation. By analogy with the Euler equation, we might expect that if r1 = r2, then the second solution contains a logarithmic term. This may also be true if the roots differ by an integer.
Equal Roots. The method of finding the second solution is essentially the same as the one we used in finding the second solution of the Euler equation (see Section 5.5) when the roots of the indicial equation were equal. We consider r to be a continuous variable and determine an as a function of r by solving the recurrence relation (8). For this choice of an (r) for n > 1, Eq. (6) reduces to
L[Ô](ã, x) = a0F(r)xr = a0(r — r1)2xr,
(15)
since r1 is a repeated root of F (r). Setting r = r1 inEq. (15), we find that L [Ô](ãð x ) = 0; hence, as we already know, y1 (x) given by Eq. (9) is one solution of Eq. (1). But more important, it also follows from Eq. (15), just as for the Euler equation, that
L
äô ä r
(rv X) = a0 — [õÃ (r — r1)2]
= a0[(r — r1) xr lnx + 2(r — r1)xr] Hence, a second solution of Eq. (1) is
0.
(16)
Ó2(õ) =
äô (r, X)
ä r
ä
r =r1 är
= (Xr1 ln X)
+ ¨ an (r )xn
n=1
Æ
0 + 1] an (r1):
n1
+xr'Y2 an (r1)x
n1
= y1(x) ln x + x^ a'n (r1)xn, x > 0,
n1
(17)
rr
rr
r
X
rr
n
a
5.7 Series Solutions near a Regular Singular Point, Part II
277
Theorem 5.7.1
where a'n (rx) denotes dan I dr evaluated at r = rv
It may turn out that it is difficult to determine an (r) as a function of r from the recurrence relation (8) and then to differentiate the resulting expression with respect to r. An alternative is simply to assume that y has the form of Eq. (17), that is,
TO
y = y1 (x) ln x + xr^Y^ bnxn, x > 0, (18)
n = 1
where y1(x) has already been found. The coefficients bn are calculated, as usual, by substituting into the differential equation, collecting terms, and setting the coefficient of each power of x equal to zero. A third possibility is to use the method of reduction of order to find y2(x) once y1(x) is known.
Roots Differing by an Integer. For this case the derivation of the second solution is considerably more complicated and will not be given here. The form of this solution is stated in Eq. (24) in the following theorem. The coefficients cn (r2) in Eq. (24) are given by
d
cn (r2) = dr [(r — r2)an (r )]
n = 1, 2,..., (19)
where an (r) is determined from the recurrence relation (8) with a0 = 1. Further, the coefficient a in Eq. (24) is
a = lim (r — r2)aN(r). (20)
r —^ 2 N
If aN(r2) is finite, then a = 0 and there is no logarithmic term in y2. A full derivation of the formulas (19), (20) may be found in Coddington (Chapter 4).
In practice the best way to determine whether a is zero in the second solution is simply to try to compute the an corresponding to the root r2 and to see whether it is possible to determine aN (r2). if so, there is no further problem. If not, we must use the form (24) with a = 0. N 2
When r1 — r2 = N, there are again three ways to find a second solution. First, we can calculate a and cn (r2) directly by substituting the expression (24) for y in Eq. (1). Second, we can calculate cn (r2) and a of Eq. (24) using the formulas (19) and (20). If this is the planned procedure, then in calculating the solution corresponding to r = r1 be sure to obtain the general formula for an (r) rather than just an (r1). The third alternative is to use the method of reduction of order.
Consider the differential equation (1),
x2 y" + x [xp(x )]y' + [x 2q (x )]y = 0,
where x = 0 is a regular singular point. Then xp(x) and x2q (x) are analytic at x = 0 with convergent power series expansions
xp(x) = pnxn, x 2q (x) = qnxn
n=0 n=0
rr
2
278
Chapter 5. Series Solutions of Second Order Linear Equations
PROBLEMS
for | x | < p, where p > 0 is the minimum of the radii of convergence of the power series for xp(x) and x2q (x). Let r1 and r2 be the roots of the indicial equation
F(r) — r (r — 1) + p0r + q0 — 0,
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