In Special Functions for Scientists and Engineers's very first section in the proof of Frobenius method, the equation (1.7)
$a_0s(s-1)+a_0q_0s+a_0r_0=0$
is obtained as a special case for $i=0$ in the general case. After some steps the general case is obtained as
$a_i[(s+i)(s+i-1)+q_0(s+i)+r_o]$ (1.11)
The next step is what confused me:
We now see that equation (1.7) reduces to
$a_0[s^2+(q_0-1)(s+r_o)]$
How can
$s(s-1)+q_0s+r_0$ $=$ $s^2+(q_0-1)(s+r_o)$ ?
Shouldn't (1.12) be
$s^2+s(q_0-1)+r_0$ ?