0 votes 0 votes The set of value(s) of $\alpha$ for which $y(t)=t^{\alpha}$ is a solution to the differential equation $$t^2 \frac{d^2y}{dx^2}-2t \frac{dy}{dx}+2y =0 \: \text{ for } t>0$$ is $\{1\}$ $\{1, -1\}$ $\{1, 2\}$ $\{-1, 2\}$ Calculus isi2016-mmamma differential-equation non-gate + – go_editor asked Sep 13, 2018 • edited Nov 19, 2019 by Lakshman Bhaiya go_editor 215 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.