#### Answer

$y=\frac{2(e^tt-e^t)+3}{e^{-t}}$

#### Work Step by Step

Since in form of $\frac{dy}{dt} + ay = g(t)$
$\mu(t) = e^{-t}$
Multiply by $\mu(t) $
$e^{-t}y'-e^{-t}y=2te^{2t}e^{-t}$
$\int{\frac{d}{dt}(e^{-t}y)} = \int{2e^tt}$
$e^{-t}y = 2(e^tt-e^t)+c$
$y=\frac{2(e^tt-e^t)+c}{e^{-t}}$
Plug in $t=0, y=1$
$1=\frac{2(-1)+c}{1}$
$c=3$