Question 11.2.7: Find two linearly independent solutions of y″ - xy′ - x²y = ...

We make the usual substitution of the power series y = \sum{c_{n} x^{n}}. This results in the equation

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

We can start the second sum at n = 0 without changing anything else. To make each term include x^{n} in its general term, we shift the index of summation in the first sum by +2 (replacen with n+2), and we shift it by -2 in the third sum (replace n with n – 2). These shifts yield

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

The common range of these three summations is n ≧ 2, so we must separate the terms corresponding to n = 0 and n = 1 in the first two sums before collecting coefficients of x^{n}. This gives

2c_{2} + 6c_{3}x – c_{1}x + \sum\limits_{n=2}^{\infty}{[(n + 2)(n + 1) c_{n+2} – nc_{n} – c_{n – 2} ]x^{n} }= 0.

The identity principle now implies that 2c_{2} = 0,  that  c_{3} = \frac{1}{6}c_{1}, and the three-term recurrence relation

c_{n+2} = \frac{nc_{n} + c_{n-2}}{(n + 2)(n + 1)}         (16)

for n ≧ 2. In particular,

c_{4} = \frac{2c_{2} + c_{0}}{12},     c_{5} = \frac{3c_{3} + c_{1}}{20},       c_{6} = \frac{4c_{4} + c_{2}}{30},

c_{7} = \frac{5c_{5} + c_{3}}{42},        c_{8} = \frac{6c_{6} + c_{4}}{56}.                            (17)

Thus all values of c_{n} for n ≧ 4 are given in terms of the arbitrary constants c_{0}  and  c_{1} because c_{2} = 0  and  c_{3} = \frac{1}{6} c_{1}. To get our first solution y_{1} of Eq. (15), we choose c_{0} = 1   and   c_{1} = 0,   so that   c_{2} = c_{3} = 0. Then the formulas in (17) yield

c_{4} = \frac{1}{12},    c_{5} = 0,   c_{6} = \frac{1}{90},    c_{7} = 0,   c_{8} = \frac{3}{1120};

thus

y_{1}(x) = 1 + \frac{1}{12}x^{4} + \frac{1}{90}x^{6} + \frac{3}{1120}x^{8} + · · ·.                    (18)

Because c_{1} = c_{3} = 0, it is clear from Eq. (16) that this series contains only terms of even degree. To obtain a second linearly independent solution y_{2} of Eq. (15), we take c_{0} = 0 and c_{1} = 1,   so   that   c_{2} = 0   and   c_{3} = \frac{1}{6} . Then the formulas in (17) yield

c_{4} = 0,       c_{5} = \frac{3}{40},      c_{6} = 0,      c_{7} =\frac{13}{1008},

so that

y_{2}(x) = x + \frac{1}{6}x^{3} + \frac{3}{40}x^{5} + \frac{13}{1008}x^{7}+ · · ·.                      (19)

Because c_{0} = c_{2} = 0, it is clear from Eq. (16) that this series contains only terms of odd degree. The solutions y_{1}(x)  and  y_{2}(x) are linearly independent because y_{1}(0) = 1 and y^{′}_{1}(0) = 0, whereas y_{2}(0) = 0 and y^{′}_{2}(0) = 1. A general solution of Eq. (15) is a linear combination of the power series in (18) and (19). Equation (15) has no singular points, so the power series representing y_{1}(x)  and    y_{2}(x) converge for all x.