Mathematical Methods for Physics and Engineering : A Comprehensive Guide

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS

is required, i.e.z= 0, is in fact an ordinary point of the ODE rather than a

regular singular point, then substitution of the Frobenius series (16.12) leads to

an indicial equation with rootsσ= 0 andσ= 1. Although these roots differ by

an integer (unity), the recurrence relations corresponding to the two roots yield

two linearly independent power series solutions (one for each root), as expected

from section 16.2.

16.4 Obtaining a second solution

Whilst attempting to construct solutions to an ODE in the form of Frobenius

series about a regular singular point, we found in the previous section that when

the indicial equation has a repeated root, or roots differing by an integer, we can

(in general) find only one solution of this form. In order to construct the general

solution to the ODE, however, we require two linearly independent solutionsy 1

andy 2. We now consider several methods for obtaining a second solution in this

case.

16.4.1 The Wronskian method

Ify 1 andy 2 are two linearly independent solutions of the standard equation

y′′+p(z)y′+q(z)y=0

then the Wronskian of these two solutions is given byW(z)=y 1 y′ 2 −y 2 y′ 1.

Dividing the Wronskian byy^21 we obtain

W

y^21

=

y 2 ′

y 1

−

y′ 1

y^21

y 2 =

y′ 2

y 1

+

[

d

dz

(

1

y 1

)]

y 2 =

d

dz

(

y 2

y 1

)

,

which integrates to give

y 2 (z)=y 1 (z)

∫z

W(u)

y 12 (u)

du.

Now using the alternative expression forW(z) given in (16.4) withC=1(since

we are not concerned with this normalising factor), we find

y 2 (z)=y 1 (z)

∫z

1

y 12 (u)

exp

{

−

∫u

p(v)dv

}

du. (16.25)

Hence, giveny 1 , we can in principle computey 2. Note that the lower limits of

integration have been omitted. If constant lower limits are included then they

merely lead to a constant times the first solution.

Find a second solution to (16.21) using the Wronskian method.

For the ODE (16.21) we havep(z)=3/(z−1), and from (16.24) we see that one solution