Find the roots of the indicial equation for $$3z^2u''+\sin(z)u'-u=0$$
My attempt:
$$z^2u''+\frac{\sin(z)}{3}u'-\frac{u}{3}=0$$
What is $P_0(z)$ and $Q_0(z)$, such that $r(r-1)+P_0r+Q_0$ is the indicial polynomial?
I'm quite certain $Q_0=-\frac{1}{3}$.
I noticed that if $P_0=\frac{1}{3}$, that is if $P_0$ is the coefficient of $zu'$ in the expansion $\frac{\sin(z)}{3}u'$, then the answer comes out.