Mathematical Methods for Physics and Engineering : A Comprehensive Guide

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS

We may check thaty 1 (z)andy 2 (z) are indeed linearly independent by computing the
Wronskian as follows:

W=y 1 y′ 2 −y 2 y′ 1

=sin

√

z

(

−

1

2

√

z

sin

√

z

)

−cos

√

z

(

1

2

√

z

cos

√

z

)

=−

1

2

√

z

(

sin^2

√

z+cos^2

√

z

)

=−

1

2

√

z

=0.

SinceW=0,thesolutionsy 1 (z)andy 2 (z) are linearly independent. Hence, the general
solution to (16.17) is given by

y(z)=c 1 sin

√

z+c 2 cos

√

z.

16.3.2 Repeated root of the indicial equation

If the indicial equation has a repeated root, so thatσ 1 =σ 2 =σ, then obviously

only one solution in the form of a Frobenius series (16.12) may be found as

described above, i.e.

y 1 (z)=zσ

∑∞

n=0

anzn.

Methods for obtaining a second, linearly independent, solution are discussed in

section 16.4.

16.3.3 Distinct roots differing by an integer

Whatever the roots of the indicial equation, the recurrence relation corresponding

to the larger of the two always leads to a solution of the ODE. However, if the

roots of the indicial equation differ by an integer then the recurrence relation

corresponding to the smaller root may or may not lead to a second linearly

independent solution, depending on the ODE under consideration. Note that for

complex roots of the indicial equation, the ‘larger’ root is taken to be the one

with the larger real part.

Find the power series solutions aboutz=0of

z(z−1)y′′+3zy′+y=0. (16.20)

Dividing through byz(z−1) to put the equation into standard form, we obtain

y′′+

3

(z−1)

y′+

1

z(z−1)

y=0, (16.21)

and on comparing with (16.7) we identifyp(z)=3/(z−1) andq(z)=1/[z(z−1)]. We
immediately see thatz= 0 is a singular point of (16.21), but sincezp(z)=3z/(z−1) and
z^2 q(z)=z/(z−1) are finite there, it is a regular singular point and we expect to find at least