Ok here's a funny ODE to solve:
y' + (x-1)y = 0)
clearly a straight forward power series substitution won't work here since we have a regular singularity at x = 0
so try the frobenius method by expanding around x = 0.
Assume
is a solution where
is some constant.
So we have
a_mx^{m+r-1})
and
(m+r-1) a_mx^{m+r-2})
Put this back in:
(m+r-1) a_mx^{m+r-2}\right) + (1-2x) \left(\sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1}\right) + (x-1) \left(\sum_{m=0}^{\infty} a_mx^{m+r}\right) = 0)
after some algebra and stuff:
^2 a_m x^{m+r-1} - \sum_{m=0}^{\infty} [2(m+r)+1] a_m x^{m+r} + \sum_{m=0}^{\infty} a_m x^{m+r+1} = 0)
clearly lowest term is
with it's coefficient as
hence
Now
, so
Now we find the coefficients of the term
where s is some constant, this gives:
^2a_{s+1} x^s - (2s+1) a_s x^s + a_{s-1} x^s = 0)
rearranging gives:
a_s - a_{s-1}}{(s+1)^2})
for s = 1, 2, etc
Thus we found a recurrence relationship with
and
as arbitrary initial values.
A bit of playing around quickly shows that:
Thus we have one of the solutions to be
x^2 + (\frac{11a_1 - 5a_0}{36})x^3 + (\frac{25a_1 - 13a_0}{288})x^4 + ...)
However because a_0 and a_1 are arbitrary, let us pick....
, now magically we have:
So
is one of the basis for the general solution of this ode.
I was wondering, since
and
are arbitrary, then would ANY
(
) and
work? Say
and
which then implies that there is an "infinite" number of different basis for the general solution of this ode? pretty cool ~
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