First notice that the given DE is a first order linear differential equation.
Rewrite it in the standard form we have
$$y'- \frac{1}{t}y = te^{-t}$$
Let $\mu(t)$ denote the integrating factor.
$$\mu(t)=e^{\int -\frac{1}{t} dt} = e^{- log(t)} = e^{log((t)^{-1})} = t^{-1}. $$
Multiply both sides of the equation by $\mu(t)$ we get
$$t^{-1}y'-t^{-2}y=e^{-t}$$
$$\frac{d}{dx}[t^{-1}y]=e^{-t}$$
Integrate both sides with respect to $t$ we get
$$t^{-1}y = -e^{-t} + C$$
So,
$$y=-te^{-t} + Ct$$
When $t \to \infty$:
Case1: if $C=0$
Then by one use of L'Hopital's rule, we get $y \to 0$.
Case2: if $C>0$
Then $y \to +\infty$.
Case3: if $C<0$
Then $y \to - \infty$.
