## frobenius method queries

Ok here's a funny ODE to solve:

$\displaystyle xy'' + (1-2x)y' + (x-1)y = 0$

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 $\displaystyle y = \sum_{m=0}^{\infty} a_mx^{m+r}$ is a solution where $\displaystyle r$ is some constant.

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

Put this back in:

$\displaystyle x \left(\sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}\right) + (1-2x) \left(\sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1}\right) + (x-1) \left(\sum_{m=0}^{\infty} a_mx^{m+r}\right) = 0$

after some algebra and stuff:

$\displaystyle \sum_{m=0}^{\infty} (m+r)^2 a_m x^{m+r-1} - \sum_{m=0}^{\infty} [2(m+r)+1] a_m x^{m+r} + \sum_{m=0}^{\infty} a_m x^{m+r+1} = 0$

clearly lowest term is $\displaystyle x^{r-1}$ with it's coefficient as$\displaystyle r^2a_0$ hence $\displaystyle r^2a_0 = 0$

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

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

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

rearranging gives:

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

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

A bit of playing around quickly shows that:

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

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

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

Thus we have one of the solutions to be $\displaystyle y = a_0 + a_1x + (\frac{3a_1 - a_0}{4})x^2 + (\frac{11a_1 - 5a_0}{36})x^3 + (\frac{25a_1 - 13a_0}{288})x^4 + ...$

However because a_0 and a_1 are arbitrary, let us pick.... $\displaystyle a_0 = a_1 = 1$, now magically we have:

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

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

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

Thanks