Answer
$y + y' = 2t - 3$.
Work Step by Step
We wish to construct a differential equation all of whose solutions are asymptotic to the line $y = 2t - 5$ as $y \rightarrow \infty$.
$Solution$:
To solve this we work "backward" from the solution(s) to the differential equation.
That is, it is obvious that the graph $$y(t) = 2t - 5 + Ce^{-t}$$ $$\rightarrow 2t - 5,$$ as $t \rightarrow \infty$, $C$, an arbitrary constant.
We can use the function $e^t$ as an integrating factor in our soon to be obtained differential equation that will yield $$y(t) = 2t - 5 + Ce^{-t}$$ as a set of solutions.
That is, the equivalent equation, with $e^t$ as an integrating factor is $$e^t y(t) = e^t(2t - 5) + C.$$
Then to obtain a differential equation that will have these as solutions, we differentiate with respect to $t$:
$$\frac{d}{dt}(e^ty(t)) = e^t(2t - 5) + e^t \cdot 2.$$
Or $$e^t y + e^t y' = e^t(2t - 5) + 2e^t.$$
Since $e^t$ is an "integrating factor", we divide through by it obtaining $$y + y' = 2t - 3$$ as our differential equation.
$Checking$:
Do all the solutions to the differential equation $y + y' = 2t - 3$ approach the line $y = 2t - 5$ as $t\rightarrow \infty$ ?
We must check that $$y(t) = 2t - 5 + Ce^{-t}$$ satisfies our differential equation, $y + y' = 2t - 3$. So, first off, differentiating our above constructed solution, we get: $$y'(t) = 2 - Ce^{-t}.$$
So then $$y + y' = (2t - 5 + Ce^{-t}) + (2 - Ce^{-t}) = 2t - 3,$$ as required. And, furthermore, $$y(t) = 2t - 5 + Ce^{-t}\rightarrow 2t - 5$$ as $t \rightarrow \infty$, as also required.
