Question 11.3.6: Find the Frobenius series solutions of xy″ + 2y′ + xy = 0. (...

In standard form the equation becomes

y″ +\frac{2}{x} y^{′} + \frac{x²}{x²}y = 0.

so we see that x = 0 is a regular singular point with p_{0} = 2  and  q_{0} = 0. The indicial equation

r(r – 1) + 2r = r(r + 1) = 0

has roots r_{1} = 0 and r_{2} = -1, which differ by an integer. In this case when r_{1} – r_{2} is an integer, it is better to depart from the standard procedure of Example 4 and begin our work with the smaller exponent. As you will see, the recurrence relation will then tell us whether or not a second Frobenius series solution exists. If it does exist, our computations will simultaneously yield both Frobenius series solutions. If the second solution does not exist, we begin anew with the larger exponent r = r_{1} to obtain the one Frobenius series solution guaranteed by Theorem 1. Hence we begin by substituting

y = x^{-1} \sum\limits_{n=0}^{\infty}{c_{n} x^{n}} = \sum\limits_{n=0}^{\infty}{c_{n} x^{n-1}}

in Eq. (31). This gives

\sum\limits_{n=0}^{\infty}{(n – 1)(n – 2)c_{n} x^{n-2} } +  2\sum\limits_{n=0}^{\infty}{(n – 1)c_{n} x^{n-2}} + \sum\limits_{n=0}^{\infty}{c_{n} x^{n} }= 0.

We combine the first two sums and shift the index by -2 in the third to obtain

\sum\limits_{n=0}^{\infty}{n(n – 1)c_{n} x^{n-2}} + \sum\limits_{n=0}^{\infty}{c_{n-2} x^{n-2}} = 0.               (32)

The cases n = 0 and n = 1 reduce to

0 · c_{0} = 0         and        0 · c_{1} = 0.

Hence we have two arbitrary constants c_{0} and c_{1} and therefore can expect to find a general solution incorporating two linearly independent Frobenius series solutions. If, for n = 1, we had obtained an equation such as 0 · c_{1} = 3, which can be satisfied for no choice of c_{1}, this would have told us that no second Frobenius series solution could exist. Now knowing that all is well, from (32) we read the recurrence relation

c_{n} = – \frac{c_{n-2}}{n(n – 1)}         for n ≧ 2.                 (33)

The first few values of n give

c_{2} = – \frac{1}{2 · 1} c_{0},                c_{3} = – \frac{1}{3 · 2}c_{1},

c_{4} = – \frac{1}{4 · 3}c_{2} = \frac{c_{0}}{4!},              c_{5} = – \frac{1}{5 · 4}c_{3} = \frac{c_{1}}{5!},

c_{6} = – \frac{1}{6 · 5}c_{4} = – \frac{c_{0}}{6!},             c_{7} = – \frac{1}{7 · 6} c_{6} = – \frac{c_{1}}{7!};

evidently the pattern is

c_{2n} = \frac{(-1)^{n} c_{0}}{(2n)!},               c_{2n+1} = \frac{(-1)^{n} c_{1}}{(2n + 1)!}

for n ≧ 1. Therefore, a general solution of Eq. (31) is

y(x) = x^{-1} \sum\limits_{n=0}^{\infty}{c_{n} x^{n}}

= \frac{c_{0}}{x}(1 – \frac{x²}{2!} + \frac{x^{4}}{4!} – · · ·) + \frac{c_{1}}{x} (x – \frac{x^{3}}{3!} + \frac{x^{5}}{5!} – · · · )

= \frac{c_{0}}{x} \sum\limits_{n=0}^{\infty}{\frac{(-1)^{n} x^{2n}}{(2n)!}} + \frac{c_{1}}{x} \sum\limits_{n=0}^{\infty}{\frac{(-1)^{n} x^{2n+1}}{(2n + 1)!}}.

Thus

y(x) = \frac{1}{x} (c_{0}  \cos x + c_{1} \sin x).

We have thus found a general solution expressed as a linear combination of the two Frobenius series solutions

y_{1}(x) = \frac{\cos x}{x}    and       y_{2}(x) = \frac{\sin x}{x}.            (34)

As indicated in Fig. 11.3.1, one of these Frobenius series solutions is bounded but the other is unbounded near the regular singular point x = 0—a common occurrence in the case of exponents differing by an integer.