Mathematical Methods for Physics and Engineering : A Comprehensive Guide

16.4 OBTAINING A SECOND SOLUTION

which is equal to zero ifσ=σ 2. As previously, since∂/∂σandLare operators

that differentiate with respect to different variables, we can reverse their order to

obtain

L

{

∂

∂σ

[(σ−σ 2 )y(z, σ)]

}

=0 atσ=σ 2 ,

and so the function
{
∂
∂σ

[(σ−σ 2 )y(z, σ)]

}

σ=σ 2

(16.30)

is also a solution of the original ODELy= 0, and is in fact the second linearly

independent solution.

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

From (16.23) the recurrence relation (withσas a parameter) is given by

(n+σ−1)an=(n+σ)an− 1.

Settinga 0 = 1 we find that the coefficients have the particularly simple forman(σ)=
(σ+n)/σ. We therefore consider the function

y(z, σ)=zσ

∑∞

n=0

an(σ)zn=zσ

∑∞

n=0

σ+n

σ

zn.

The smaller root of the indicial equation for (16.21) isσ 2 = 0, and so from (16.30) a
second, linearly independent, solution to the ODE is given by
{
∂
∂σ

[σy(z, σ)]

}

σ=0

=

{

∂

∂σ

[

zσ

∑∞

n=0

(σ+n)zn

]}

σ=0

.

The derivative with respect toσis given by

∂

∂σ

[

zσ

∑∞

n=0

(σ+n)zn

]

=zσlnz

∑∞

n=0

(σ+n)zn+zσ

∑∞

n=0

zn,

which on settingσ= 0 gives the second solution

y 2 (z)=lnz

∑∞

n=0

nzn+

∑∞

n=0

zn

=

z

(1−z)^2

lnz+

1

1 −z

=

z

(1−z)^2

(

lnz+

1

z

− 1

)

.

This second solution is the same as that obtained by the Wronskian method in the previous
subsection except for the addition of some of the first solution.

16.4.3 Series form of the second solution

Using any of the methods discussed above, we can find the general form of the

second solution to the ODE. This form is most easily found, however, using the