16.4 OBTAINING A SECOND SOLUTION
to (16.21) isy 1 =z/(1−z)^2. Substituting forpandy 1 in (16.25) we have
y 2 (z)=z
(1−z)^2∫z
(1−u)^4
u^2exp(
−
∫u
3
v− 1dv)
du=
z
(1−z)^2∫z
(1−u)^4
u^2exp[−3ln(u−1)]du=
z
(1−z)^2∫z
u− 1
u^2du=
z
(1−z)^2(
lnz+1
z)
.
By calculating the Wronskian ofy 1 andy 2 it is easily shown that, as expected, the two
solutions are linearly independent. In fact, as the Wronskian has already been evaluated
asW(u)=exp[−3ln(u−1)], i.e.W(z)=(z−1)−^3 , no calculation is needed.
An alternative (but equivalent) method of finding a second solution is simply toassume that the second solution has the formy 2 (z)=u(z)y 1 (z) for some function
u(z) to be determined (this method was discussed more fully in subsection 15.2.3).
From (16.25), we see that the second solution derived from the Wronskian is
indeed of this form. Substitutingy 2 (z)=u(z)y 1 (z) into the ODE leads to a
first-order ODE in whichu′is the dependent variable; this may then be solved.
16.4.2 The derivative methodThe derivative method of finding a second solution begins with the derivation of
a recurrence relation for the coefficientsanin a Frobenius series solution, as in the
previous section. However, rather than puttingσ=σ 1 in this recurrence relation
to evaluate the first series solution, we now keepσas a variable parameter. This
means that the computedanare functions ofσand the computed solution is now
a function ofzandσ:
y(z, σ)=zσ∑∞n=0an(σ)zn. (16.26)Of course, if we putσ=σ 1 in this, we obtain immediately the first series solution,
but for the moment we leaveσas a parameter.
For brevity let us denote the differential operator on the LHS of our standardODE (16.7) byL,sothat
L=d^2
dz^2+p(z)d
dz+q(z),and examine the effect ofLon the seriesy(z, σ) in (16.26). It is clear that the
seriesLy(z, σ) will contain only a term inzσ, since the recurrence relation defining
thean(σ) is such that these coefficients vanish for higher powers ofz.Butthe
coefficient ofzσis simply the LHS of the indicial equation. Therefore, if the roots