I tried solving $2xy''-(3+2x)y'+y=0$ using the Frobenius method, and I kept getting incorrect answers.
I got a final form which is
$$a_{k+1} = \frac{(-2k-2c+1)a_k }{ 2k^2+2c^2+4kc-c-k-3}$$
this when I input $k=0$ I got $a_1= \frac{-1}3 a_0$
but the answer key instead does this
\begin{align} (2r^2-5r)C_0x^{r-1}+\sum_{k=1}^\infty\Bigl(2(k+r)(k+r-1)C_k-3(k+r)C_k&\\ -2(k+r-1)C_{k-1}+C_{k-1}\Bigr)x^{r+k-1}&=0 \end{align} original formula
There the only difference is that it sets the equation to $x^{r+k-1}$ while I replaced $n=1$ with $n=k+1$ and ended up with $x^{c+k}$ ( the answer key uses $r$ instead of $c$) I don't get the reasons for doing this either because essentially I thought it wouldn't make a difference by solving it either way but I just can't figure out why I keep getting the incorrect answer.
I have checked for calculation error multiple times or am I missing something?
apologies if I made my point confusing all I need to know is that when we round the equation to $\sigma k=0$ is it solvable? and it would be great if I can see the process to see if I did anything incorrectly.
The first equation you have to consider is for $k=-1$, with $a_{-1}=0$. To get a non-trivial $a_0$, you need that the denominator is zero, which is also called the indicial equation. The denominator for $k=-1$ is $$ 2+2c^2-4c-c+1-3=2c^2-5c, $$ which is the same as the first coefficient in front of $C_0$ in the solution key.
Thus you get one basis solution in the power series for $c=0$ and another one in the power series for $c=\frac52$.