Mathematical Methods for Physics and Engineering : A Comprehensive Guide

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS

power series of the form (see section 24.11)

y(z)=

∑∞

n=0

anzn. (16.9)

Moreover, it may be shown that such a power series converges for|z|<R,where

Ris the radius of convergence and is equal to the distance fromz=0tothe

nearest singular point of the ODE (see chapter 24). At the radius of convergence,

however, the series may or may not converge (as shown in section 4.5).

Since every solution of (16.7) is analytic at an ordinary point, it is always

possible to obtain twoindependentsolutions (from which the general solution

(16.2) can be constructed) of the form (16.9). The derivatives ofywith respect to

zare given by

y′=

∑∞

n=0

nanzn−^1 =

∑∞

n=0

(n+1)an+1zn, (16.10)

y′′=

∑∞

n=0

n(n−1)anzn−^2 =

∑∞

n=0

(n+2)(n+1)an+2zn. (16.11)

Note that, in each case, in the first equality the sum can still start atn=0since

the first term in (16.10) and the first two terms in (16.11) are automatically zero.

The second equality in each case is obtained by shifting the summation index

so that the sum can be written in terms of coefficients ofzn. By substituting

(16.9)–(16.11) into the ODE (16.7), and requiring that the coefficients of each

power ofzsum to zero, we obtain arecurrence relationexpressing eachanin

terms of the previousar(0≤r≤n−1).

Find the series solutions, aboutz=0,of

y′′+y=0.

By inspection,z= 0 is an ordinary point of the equation, and so we may obtain two
independent solutions by making the substitutiony=

∑∞

n=0anz

n. Using (16.9) and (16.11)

we find
∑∞

n=0

(n+2)(n+1)an+2zn+

∑∞

n=0

anzn=0,

which may be written as

∑∞

n=0

[(n+2)(n+1)an+2+an]zn=0.

For this equation to be satisfied we require that the coefficient of each power ofzvanishes
separately, and so we obtain the two-term recurrence relation

an+2=−

an

(n+2)(n+1)

forn≥ 0.

Using this relation, we can calculate, say, the even coefficientsa 2 ,a 4 ,a 6 andsoon,for