SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
of the indicial equation areσ=σ 1 andσ=σ 2 then it follows that
Ly(z, σ)=a 0 (σ−σ 1 )(σ−σ 2 )zσ. (16.27)
Therefore, as in the previous section, we see that fory(z, σ)tobeasolutionof
the ODELy=0,σmust equalσ 1 orσ 2. For simplicity we shall seta 0 = 1 in the
following discussion.
Let us first consider the case in which the two roots of the indicial equation
are equal, i.e.σ 2 =σ 1. From (16.27) we then have
Ly(z, σ)=(σ−σ 1 )^2 zσ.
Differentiating this equation with respect toσwe obtain
∂
∂σ
[Ly(z, σ)]=(σ−σ 1 )^2 zσlnz+2(σ−σ 1 )zσ,
which equals zero ifσ=σ 1. But since∂/∂σandLare operators that differentiate
with respect to different variables, we can reverse their order, implying that
L
[
∂
∂σ
y(z, σ)
]
=0 atσ=σ 1.
Hence, the function in square brackets, evaluated atσ=σ 1 and denoted by
[
∂
∂σ
y(z, σ)
]
σ=σ 1
, (16.28)
is also a solution of the original ODELy= 0, and is in fact the second linearly
independent solution that we were looking for.
The case in which the roots of the indicial equation differ by an integer is
slightly more complicated but can be treated in a similar way. In (16.27), sinceL
differentiates with respect tozwe may multiply (16.27) by any function ofσ,say
σ−σ 2 , and take this function inside the operatorLon the LHS to obtain
L[(σ−σ 2 )y(z, σ)]=(σ−σ 1 )(σ−σ 2 )^2 zσ. (16.29)
Therefore the function
[(σ−σ 2 )y(z, σ)]σ=σ 2
is also a solution of the ODELy= 0. However, it can be proved§that this
function is a simple multiple of the first solutiony(z, σ 1 ), showing that it is not
linearly independent and that we must find another solution. To do this we
differentiate (16.29) with respect toσand find
∂
∂σ
{L[(σ−σ 2 )y(z, σ)]}=(σ−σ 2 )^2 zσ+2(σ−σ 1 )(σ−σ 2 )zσ
+(σ−σ 1 )(σ−σ 2 )^2 zσlnz,
§For a fuller discussion see, for example, K. F. Riley,Mathematical Methods for the Physical Sciences
(Cambridge: Cambridge University Press, 1974), pp. 158–9.