Mathematical Methods for Physics and Engineering : A Comprehensive Guide

16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINT

agivena 0. Alternatively, starting witha 1 , we obtain the odd coefficientsa 3 ,a 5 ,etc.Two
independent solutions of the ODE can be obtained by setting eithera 0 =0ora 1 =0.
Firstly, if we seta 1 = 0 and choosea 0 = 1 then we obtain the solution

y 1 (z)=1−

z^2

2!

+

z^4

4!

−···=

∑∞

n=0

(−1)n

(2n)!

z^2 n.

Secondly, if we seta 0 = 0 and choosea 1 = 1 then we obtain a second,independent, solution

y 2 (z)=z−

z^3

3!

+

z^5

5!

−···=

∑∞

n=0

(−1)n

(2n+1)!

z^2 n+1.

Recognising these two series as coszand sinz, we can write the general solution as

y(z)=c 1 cosz+c 2 sinz,

wherec 1 andc 2 are arbitrary constants that are fixed by boundary conditions (if supplied).
We note that both solutions converge for allz, as might be expected since the ODE
possesses no singular points (except|z|→∞).

Solving the above example was quite straightforward and the resulting series

were easily recognised and written inclosed form(i.e. in terms of elementary

functions);this is not usually the case. Another simplifying feature of the previous

example was that we obtained a two-term recurrence relation relatingan+2and

an, so that the odd- and even-numbered coefficients were independent of one

another. In general, the recurrence relation expressesanin terms of any number

of the previousar(0≤r≤n−1).

Find the series solutions, aboutz=0,of

y′′−

2

(1−z)^2

y=0.

By inspection,z= 0 is an ordinary point, and therefore we may find two independent
solutions by substitutingy=

∑∞

n=0anz

n. Using (16.10) and (16.11), and multiplying through

by (1−z)^2 , we find

(1− 2 z+z^2 )

∑∞

n=0

n(n−1)anzn−^2 − 2

∑∞

n=0

anzn=0,

which leads to

∑∞

n=0

n(n−1)anzn−^2 − 2

∑∞

n=0

n(n−1)anzn−^1 +

∑∞

n=0

n(n−1)anzn− 2

∑∞

n=0

anzn=0.

In order to write all these series in terms of the coefficients ofzn, we must shift the
summation index in the first two sums, obtaining

∑∞

n=0

(n+2)(n+1)an+2zn− 2

∑∞

n=0

(n+1)nan+1zn+

∑∞

n=0

(n^2 −n−2)anzn=0,

which can be written as

∑∞

n=0

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