Determine the value of r for which the given differential equation has solutions of the form $\displaystyle y=t^r$ for t > 0.

$\displaystyle t^2 y'' + 4ty' + 2y = 0$

$\displaystyle y'(t) = rt^{r-1}$

$\displaystyle y''(t) = r(r-1)t^{r-2}$

$\displaystyle t^2[r(r-1)t^{r-2}] + 4t[rt^{r-1}] + 2t^r = 0$

I'm a little confused as how to simplify this down. Can $\displaystyle t^2$ multiplied by $\displaystyle t^{r-2}$ produce $\displaystyle t^r$ ?