Constructing a linear first order ODE with convergent solutions. I am studying for a test and cannot figure out for the life of me how to do this problem. I need to construct a first order linear ODE in the form of $y'+p(t)y=g(t)$ such that all of the solutions of the ODE satisfy the condition that the limit as $t$ approaches infinity, $y(t)=4-t^2$.
Now I know one such solution to be $y'+(y/t)=(4/t)-3t$. I understand why that is a solution, because the general solution to the ODE is $y(t)=4-t^2+e/t$.
What I don't understand is how they came to that solution in the first place. How does one work backwards to find the ODE?

Your condition on the solution is not that clear, as y(t) is clearly different from 4-t² near infinity. Close, sure, but different. So the question is : does it means that y(t)-(4-t^2) converge to 0, or that y(t)/(4-t^2) converge to 1, or another asymptotic "closeness" definition.
Assuming that it means that  $y(t)-(4-t^2)$ converge to 0, you get that $y(t) = (4-t^2)+\epsilon(t)$, with $\epsilon(t)$ that converge to $0$ at infinity
You plug it into the equation and you get :
$$-2t + \epsilon'(t) + p(t)(4-t^2) + p(t) \epsilon(t) = g(t)$$
$$ \epsilon'(t) +  p(t) \epsilon(t)  = g(t) - p(t)(4-t^2) + 2t$$
Then you say : what if $p(t) = 1$ and $g(t) - p(t)(4-t^2) + 2t = 0$, we would then have $\epsilon(t) = Ce^{-t}$ that converge indeed to 0
This is of course one possibility, and there can be many others. Note that it works not only for 4-t^2, but for any derivable function (assuming this particular definition of "closeness")

Now what if I find p(t) = 1 boring? How to spice things a little, while still getting an $\epsilon(t)$ that converge to 0?
Simple : take a function A(t) that is derivable with $\lim_{t\to +\infty} A(t) = - \infty$, and choose $p(t) = -A'(t)$ and $g(t) = -A'(t)(4−t^2)+2t$
Now the equation that verify $\epsilon$ become 
$$\epsilon'(t) - A'(t) \epsilon(t) = 0$$
Hence $\epsilon(t) = C \exp(A(t) )$, and by definition of $A$, the limit of epsilon is indeed 0.
