Ok here's a funny ODE to solve:

xy'' + (1-2x)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 y = \sum_{m=0}^{\infty} a_mx^{m+r} is a solution where r is some constant.

So we have y' = \sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1} and y'' = \sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}

Put this back in:

x \left(\sum_{m=0}^{\infty}(m+r)(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:

\sum_{m=0}^{\infty} (m+r)^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 x^{r-1} with it's coefficient as  r^2a_0 hence r^2a_0 = 0

Now a_0 \neq 0, so r^2 = 0 \implies r = 0

Now we find the coefficients of the term x^s where s is some constant, this gives:

(s+1)^2a_{s+1} x^s - (2s+1) a_s x^s + a_{s-1} x^s = 0

rearranging gives:

a_{s+1} = \frac{(2s+1)a_s - a_{s-1}}{(s+1)^2} for s = 1, 2, etc

Thus we found a recurrence relationship with a_0 and a_1 as arbitrary initial values.

A bit of playing around quickly shows that:

a_2 = \frac{3a_1 - a_0}{4}

a_3 = \frac{11a_1 - 5a_0}{36}

a_4 = \frac{25a_1 - 13a_0}{288}

Thus we have one of the solutions to be y = a_0 + a_1x + (\frac{3a_1 - a_0}{4})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.... a_0 = a_1 = 1 , now magically we have:

y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+ ... = e^x

So y = e^x is one of the basis for the general solution of this ode.

Now I was wondering, since a_0 and a_1 are arbitrary, then would ANY a_0 ( \neq 0) and a_1 work? Say a_0 = 4 and a_1 = 3 which then implies that there is an "infinite" number of different basis for the general solution of this ode?

Thanks