Question

Need help with indicial equation and ODEs

Answer 1

With

$y=\displaystyle\sum_{n\ge0}a_nx^{n+r}$
$y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}$
$y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}$

the singular ODE can be written as

$\displaystyle\sum_{n\ge0}\bigg[2(n+r)(n+r-1)+3(n+r)-1\bigg]a_nx^{n+r}+\sum_{n\ge0}\bigg[-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^{n+r-1}=0$

The first term of the second series admits the indicial equation. When $n=0$, we have

$(-2r(r-1)-3r)a_0x^{r-1}=0\iff-2r^2-r=0\iffr^2+\dfrac12r=0$

Factoring reveals two distinct roots at $-r(2r+1)=0\implies r_1=0,r_2=-\dfrac12$ (in your case, swap $r_1$ and $r_2$ before submitting).

Next, shift the index of the first sum so that it starts at $n=1$ by replacing $n\mapsto n-1$, then consolidate the sums to get

$\displaystyle\sum_{n\ge1}\bigg[2(n+r-1)(n+r-2)+3(n+r-1)-1-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^{n+r-1}$
$=\displaystyle x^r\sum_{n\ge1}\bigg[-4n-4r\bigg]a_nx^{n-1}$

Setting $r=-\dfrac12$, we then have this as

$=x^{-1/2}\left(-2a_1-6a_2x-10a_3x^2-14a_4x^3-18a_5x^4+\cdots\right)$
$=x^{-1/2}\displaystyle\sum_{n\ge1}(2-4n)a_nx^{n-1}$

However I don't see the connection to the given answer... It seems some information is missing, specifically about how the coefficients $a_n$ are related.