SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
derivative method. Let us first consider the case where the two solutions of the
indicial equation are equal. In this case a second solution is given by (16.28),
which may be written as
y 2 (z)=[
∂y(z, σ)
∂σ]σ=σ 1=(lnz)zσ^1∑∞n=0an(σ 1 )zn+zσ^1∑∞n=1[
dan(σ)
dσ]σ=σ 1zn=y 1 (z)lnz+zσ^1∑∞n=1bnzn, (16.31)wherebn=[dan(σ)/dσ]σ=σ 1. One could equally obtain the coefficientsbnby direct
substitution of the form (16.31) into the original ODE.
In the case where the roots of the indicial equation differ by an integer (notequal to zero), then from (16.30) a second solution is given by
y 2 (z)={
∂
∂σ[(σ−σ 2 )y(z, σ)]}σ=σ 2=lnz[(σ−σ 2 )zσ∑∞n=0an(σ)zn]σ=σ 2+zσ^2∑∞n=0[
d
dσ(σ−σ 2 )an(σ)]σ=σ 2zn.But, as we mentioned in the previous section,[(σ−σ 2 )y(z, σ)]atσ=σ 2 is just a
multiple of the first solutiony(z, σ 1 ). Therefore the second solution is of the form
y 2 (z)=cy 1 (z)lnz+zσ^2∑∞n=0bnzn, (16.32)wherecis a constant. In some cases, however,cmight be zero, and so the second
solution would not contain the term in lnzand could be written simply as a
Frobenius series. Clearly this corresponds to the case in which the substitution
of a Frobenius series into the original ODE yields two solutions automatically.
In either case, the coefficientsbnmay also be found by direct substitution of the
form (16.32) into the original ODE.
16.5 Polynomial solutionsWe have seen that the evaluation of successive terms of a series solution to a
differential equation is carried out by means of a recurrence relation. The form
of the relation forandepends uponn, the previous values ofar(r<n)andthe
parameters of the equation. It may happen, as a result of this, that for some
value ofn=N+ 1 the computed valueaN+1is zero and that all higheraralso
vanish. If this is so, and the corresponding solution of the indicial equationσ