Answer
$y + y' = 2 - t$
Work Step by Step
We wish to construct a first order linear differential equation all of whose solutions are asymptotic to the line $y = 3 - t$ as $t \rightarrow \infty$.
SOLUTION.
After a few trials we hit upon
(1) $y(t) = 3 - t + Ce^{-t} \rightarrow 3 - t$ as $t \rightarrow \infty$.
So if we consider $e^t$ as an integrating factor for our sought after differential equation, we would then have
(2) $e^t y(t) = e^t (3 - t) + C$.
We can then differentiate both sides of (2) getting
$e^t y + e^t y' = e^t(3 - t) + e^t (-1)$,
which, after factoring out $e^t$, will give us our desired differential equation:
(3) $y + y' = 2 - t$
with $e^t$ being an integrating factor.
CHECK.
Does $y = 3 - t + Ce^{-t}$ satisfy $y + y' = 2 - t$?
Differentiating our expression for $y$,
$y' = -1 - Ce^{-t}$.
Substituting our expressions for $y$ and $y'$ into our proposed equation (3), we get
$(3 - t + Ce^{-t}) + (-1 - Ce^{-t}) = 2 -t$.
I.e., $y + y' = 2 - t$,
and, by our initial construction (1), we see that $y(t) \rightarrow 3 - t$ as $t \rightarrow \infty$, as required.
