Answer
$y + y' = 4 - t^2 -2t$
Work Step by Step
We wish to find a first order linear differential equation all of whose solutions approach the curve
$$y = 4 - t^2$$ as $$t \rightarrow \infty .$$
$Solution$: We will "work backward" in the sense of elucidating a family of functions, depending on $t$, all of whose graphs approach the graph $y = 4 - t^2$ as $t \rightarrow \infty$, and then, by differentiation, obtain a differential equation for which this family of functions are its solutions.
Noting that $Ce^{-t} \rightarrow 0$ as $t \rightarrow \infty$, where $C$ is an arbitrary constant, it is then immediate that
$$y(t) = 4 - t^2 - Ce^{-t} \rightarrow 4 - t^2$$ as $$t \rightarrow \infty.$$
So to get a differential equation which will have these functions, $y(t)$, (each such function having a different real number value for $C$), we can use the function $e^t$ as an integrating factor for our yet to be obtained differential equation. That is, we can write, as an equivalent functional equation,
$$e^t y = e^t (4 - t^2) + C.$$
Now differentiating to get the required differential equation:
$$\frac{d}{dt}(e^t y) = \frac{d}{dt}[e^t(4 - t^2) + C],$$
$$e^t y + e^t y' = e^t(4 - t^2) + e^t\cdot (-2t),$$
$$y + y' = 4 - t^2 -2t,$$
which will be our differential equation.
$Checking$: We already know from our initial construction, above, that a function of the form $y(t) = 4 - t^2 + Ce^{-t}$ approaches the graph $y = 4 - t^2$ as $t$ approaches $\infty$.
We then need only to show that such a function satisfies our obtained differential equation, $y + y' = 4 - t^2 -2t$. Toward that end we differentiate such a function and get $y'(t) = -2t - Ce^{-t}$.
So substituting such a function and its derivative into our obtained differential equation we get,
$$(4 - t^2 + Ce^{-t}) + (-2t - Ce^{-t}) = 4 - t^2 - 2t,$$
which was to be shown.
