Answer
$y=\frac{t^2-1}{2e^{2t}}$
Work Step by Step
Since in form of $\frac{dy}{dt}+ay=g(t)$
$\mu(t)=e^{2t}$
$e^{2t}y'+e^{2t}2y=e^{2t}e^{-2t}t$
$\frac{d}{dt}(e^{2t}y)=e^{-2t}e^{2t}t$
$\int{\frac{d}{dt}(e^{2t}y)}=\int{e^{0}t}$
$e^{2t}y=\frac{t^2}{2}+\frac{2c}{2}$
Constant 2c/2 can be simplified to c, since it can be any constant.
$y=\frac{t^2+c}{2e^{2t}}$
Plug in $y=0, t=1$
$0=\frac{1+c}{2e^2}$
$c=-1$
