Let
[tex]y = t^r \\ \\ y' = rt^{r-1} \\ \\ y'' = r(r-1) t^{r-2}[/tex]
Substitute into differential equation:
[tex](r^2 -r) t^r - (5r) t^r + (5) t^r = 0[/tex]
Factor out t^r and solve for 'r':
[tex]r^2 - r -5r +5 = 0 \\ \\ r^2 - 6r+5 = 0 \\ \\ (r-5)(r-1) = 0 \\ \\ r = 1, r=5[/tex]