Question 11.3.4: Find the Frobenius series solutions of 2x²y″ + 3xy′ - (x² + ...

First we divide each term by 2x² to put the equation in the form in (17):

y^{″} + \frac{p_{0} + p_{1} x + p_{2}x² + . . .}{x}y^{′} + \frac{q_{0} +  q_{1} x + q_{2}x² + . . .}{x²} y = 0.           (17)

y^{″} + \frac{\frac{3}{2} }{x} y^{′} + \frac{- \frac{1}{2} – \frac{1}{2} x²}{x²}y = 0.                                       (21)

We now see that x = 0 is a regular singular point, and that p_{0} = \frac{3}{2}   and   q_{0} = – \frac{1}{2} . Because p(x) ≡ \frac{3}{2}   and   q(x) = -\frac{1}{2} – \frac{1}{2} x² are polynomials, the Frobenius series we obtain will converge for all x > 0. The indicial equation is

r(r – 1) + \frac{3}{2} r – \frac{1}{2} = (r – \frac{1}{2})(r + 1) = 0,

so the exponents are r_{1} = \frac{1}{2}  and  r_{2} = -1. They do not differ by an integer, so Theorem 1 guarantees the existence of two linearly independent Frobenius series solutions. Rather than separately substituting

y_{1} = x^{1/2} \sum\limits_{n =1}^{\infty}{a^{n} x^{n}}     and    y_{2} = x^{-1} \sum\limits_{n =1}^{\infty}{b_{n} x^{n}}

in Eq. (20), it is more efficient to begin by substituting y = x^{r} ∑ c_{n} x^{n}. We will then get a recurrence relation that depends on r. With the value r_{1} = \frac{1}{2} it becomes a recurrence relation for the series for y_{1}, whereas with r_{2} = – 1 it becomes a recurrence relation for the series for y_{2}. When we substitute

y = \sum\limits_{n=0}^{\infty}{c_{n} x^{n + r}},        y^{′} = \sum\limits_{n=0}^{\infty}{(n + r) c_{n} x^{n + r – 1}},

and

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

in Eq. (20)—the original differential equation, rather than Eq. (21)—we get

2 \sum\limits_{n=0}^{\infty}{(n + r)(n + r – 1)c_{n} x^{n+r} }+ 3 \sum\limits_{n=0}^{\infty}{(n + r)c_{n} x^{n+r} } – \sum\limits_{n=0}^{\infty}{c_{n} x^{n+r+2}} – \sum\limits_{n=0}^{\infty}{c_{n} x^{n+r }}= 0.        (22)

At this stage there are several ways to proceed. A good standard practice is to shift indices so that each exponent will be the same as the smallest one present. In this example, we shift the index of summation in the third sum by -2 to reduce its exponent from n + r + 2 to n + r. This gives

2 \sum\limits_{n=0}^{\infty}{(n + r)(n + r – 1)c_{n} x^{n+r} }+ 3 \sum\limits_{n=0}^{\infty}{(n + r)c_{n} x^{n+r} } –  \sum\limits_{n=2}^{\infty}{c_{n-2} x^{n+r}} – \sum\limits_{n=0}^{\infty}{c_{n} x^{n+r }}= 0.         (23)

The common range of summation is n\geqq  2, so we must treat n = 0 and n = 1 separately. Following our standard practice, the terms corresponding to n = 0 will always give the indicial equation

[2r(r – 1) + 3r – 1]c_{0} = 2 (r^{2} + \frac{1}{2}r – \frac{1}{2})c_{0} = 0.

The terms corresponding to n = 1 yield

[2(r + 1)r + 3(r + 1) – 1]c_{1} = (2r^{2} + 5r + 2)c_{1} = 0.

Because the coefficient 2r² + 5r + 2 of c_{1} is nonzero whether r = \frac{1}{2}  or  r = – 1, it follows that

c_{1} = 0                            (24)

in either case.
The coefficient of x^{n+r} in Eq. (23) is

2(n + r)(n + r – 1)c_{n} + 3(n + r)c_{n} – c_{n-2} – c_{n} = 0.

We solve for c_{n} and simplify to obtain the recurrence relation

c_{n} = \frac{c_{n-2}}{2(n + r)^{2} + (n + r) – 1}    for n ≧ 2.        (25)

CASE 1: r_{1} = \frac{1}{2} . We now write a_{n} in place of c_{n} and substitute r = \frac{1}{2} in Eq. (25). This gives the recurrence relation

a_{n} = \frac{a_{n-2}}{2n² + 3n}       for n ≧ 2.               (26)

With this formula we can determine the coefficients in the first Frobenius solution y_{1}. In view of Eq. (24) we see that a_{n} = 0 whenever n is odd. With n = 2, 4, and 6 in Eq. (26), we get

a_{2} = \frac{a_{0}}{14},     a_{4} = \frac{a_{2}}{44} = \frac{a_{0}}{616} ,           and        a_{6} = \frac{a_{4}}{90} = \frac{a_{0}}{55,440}.

Hence the first Frobenius solution is

y_{1}(x) = a_{0}x^{1/2} (1 + \frac{x^{2}}{14} + \frac{x^{4}}{616} + \frac{x^{6}}{55,440}+ . . . ).

CASE 2: r_{2} = -1. We now write b_{n} in place of c_{n} and substitute r = -1 in Eq. (25). This gives the recurrence relation

b_{n} = \frac{b_{n-2}}{2b² – 3n}       for n ≧ 2.   (27)

Again, Eq. (24) implies that b_{n} = 0 for n odd. With n = 2, 4, and 6 in (27), we get

b_{2} = \frac{b_{0}}{2},     b_{4} = \frac{b_{2}}{20} = \frac{b_{0}}{40} ,           and       b_{6} = \frac{b_{4}}{54} = \frac{b_{0}}{2160}.

Hence the second Frobenius solution is

y_{2}(x) = b_{0}x^{-1} (1 + \frac{x^{2}}{2} + \frac{x^{4}}{40} + \frac{x^{6}}{2160}+ . . . ).