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[tex]xy'' + (1-2x)y' + (x-1)y = 0[/tex]

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 [tex]y = \sum_{m=0}^{\infty} a_mx^{m+r}[/tex] is a solution where [tex]r[/tex] is some constant.

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

Put this back in:

[tex]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[/tex]

after some algebra and stuff:

[tex]\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[/tex]

clearly lowest term is [tex]x^{r-1}[/tex] with it's coefficient as[tex] r^2a_0[/tex] hence [tex]r^2a_0 = 0[/tex]

Now [tex]a_0 \neq 0[/tex], so [tex]r^2 = 0 \implies r = 0[/tex]

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

[tex](s+1)^2a_{s+1} x^s - (2s+1) a_s x^s + a_{s-1} x^s = 0[/tex]

rearranging gives:

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

Thus we found a recurrence relationship with [tex]a_0[/tex] and [tex]a_1[/tex] as arbitrary initial values.

A bit of playing around quickly shows that:

[tex]a_2 = \frac{3a_1 - a_0}{4}[/tex]

[tex]a_3 = \frac{11a_1 - 5a_0}{36}[/tex]

[tex]a_4 = \frac{25a_1 - 13a_0}{288}[/tex]

Thus we have one of the solutions to be [tex]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 + ...[/tex]

However because a_0 and a_1 are arbitrary, let us pick.... [tex]a_0 = a_1 = 1 [/tex], now magically we have:

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

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

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

Thanks