0 votes 0 votes The solution of the differential equation, for $t > 0, y"(t)+2y'(t)+y(t)=0$ with initial conditions $y(0)=0$ and $y'(0)=1$, is ($u(t)$ denotes the unit step function), $te^{-t}u(t)$ $(e^{-t}-te^{-t})u(t)$ $(-e^{-t}+te^{-t})u(t)$ $e^{-t}u(t)$ Differential Equations gate2016-ee-2 differential-equations + – makhdoom ghaya asked Jan 29, 2017 retagged Mar 9, 2021 by Lakshman Bhaiya makhdoom ghaya 9.4k points answer comment Share See all 0 reply Please log in or register to add a comment.