Page 299 - Schaum's Outline of Differential Equations

282 SERIES SOLUTIONS NEAR A REGULAR SINGULAR POINT [CHAP. 28

z
28.12. Use the method of Frobenius to find one solution near x = 0 of x y" - xy' + y = 0.
Here P(x) = -1/x and Q(x) = l/x , 2 so x = 0 is a regular singular point and the method of Frobenius is applicable.
Substituting Eqs. (28.2) through (28.4) into the left side of the differential equation, as given, and combining coefficients
of like powers of x, we obtain

Thus,
and, in general,

2
From (1), the indicial equation is (A - I) = 0, which has roots A : = A 2 = 1. Substituting X = 1 into (2), we obtain
2
n a n = 0, which implies that a n = 0, n > 1. Thus, yi(x) = a^.
28.13. Find the general solution near x = 0 to the differential equation given in Problem 28.12.
One solution is given in Problem 28.12. Because the roots of the indicial equation are equal, we use Eq. (28.8)
to generate a second linearly independent solution. The recurrence formula is (2) of Problem 28.12. Solving it for
a n, in terms of A, we find that a n = Q (n> 1), and when these values are substituted into Eq. (28.2), we have
y (A, x) = a 0x^. Thus,

and

which is precisely the form of Eq. (28.7), where, for this particular differential equation, b n(ki) = 0(n = 0, 1,2, ...).
The general solution is

where fcj = C^Q, and k 2 = C 2a 0.

28.14. Use the method of Frobenius to find one solution near x = 0 of x^y" + (x 2 - 2x)y' + 2y = 0.
Here

so x = 0 is a regular singular point and the method of Frobenius is applicable. Substituting, Eqs. (28.2) through
(28.4) into the left side of the differential equation, as given, and combining coefficients of like powers of x, we
obtain

Dividing by x^, factoring the coefficient of a n, and equating the coefficient of each power of x to zero, we obtain

and, in general, [(A + n) - 2] [(A + n) - l]a n + (A + n - l)a n _ i = 0, or,

2
From (_/), the indicial equation is A - 3A + 2 = 0, which has roots A : = 2 and A 2 = 1. Since A : - A^ = 1, a positive
integer, the solution is given by Eqs. (28.5) and (28.9). Substituting A, = 2 into (2), we have a n = —(lln)a n_ 1,