Question 11.1.2: Solve the equation (x - 3)y′ + 2y = 0.

As before, we substitute

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

to obtain

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

so that

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

In the first sum we can replace n = 1 with n = 0 with no effect on the sum. In the second sum we shift the index of summation by +1. This yields

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

that is,

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

The identity principle then gives

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

from which we obtain the recurrence relation

c_{n+1} = \frac{n + 2}{3(n + 1)} c_{n}        for n ≧ 0.

We apply this formula with n = 0, n = 1, and n = 2, in turn, and find that

c_{1} = \frac{2}{3}c_{0},      c_{2} = \frac{3}{3 · 2}c_{1} = \frac{3}{3²}c_{0},          and     c_{3} = \frac{4}{3 · 3}c_{2} = \frac{4}{3³}c_{0}.

This is almost enough to make the pattern evident; it is not difficult to show by induction on n that

c_{n} = \frac{n + 1}{3^{n}}c_{0}      if n ≧ 1.

Hence our proposed power series solution is

y(x) = c_{0} \sum\limits_{n=0}^{\infty}{\frac{n + 1}{3^{n}}x^{n}}.                                  (26)

Its radius of convergence is

ρ = \underset{n → \infty }{\lim} \left| \frac{c_{n}}{c_{n} + 1} \right|  = \underset{n → \infty }{\lim} \frac{3n + 3}{n + 2} = 3.

Thus the series in (26) converges if -3 < x < 3 and diverges if |x| > 3. In this particular example we can explain why. An elementary solution (obtained by separation of variables) of our differential equation is y = 1/(3 – x)². If we differentiate termwise the geometric series

\frac{1}{3 – x} = \frac{\frac{1}{3}}{1 – \frac{x}{3}} = \frac{1}{3} \sum\limits_{n=0}^{\infty}{\frac{x^{n}}{3^{n}}},

we get a constant multiple of the series in (26). Thus this series (with the arbitrary constant c_{0} appropriately chosen) represents the solution

y(x) = \frac{1}{(3 – x)^{2}}

on the interval -3 < x < 3, and the singularity at x = 3 is the reason why the radius of convergence of the power series solution turned out to be ρ = 3.