Equilibrium Solution in Solving a Linear ODE using an Integrating Factor Problem
I've stumbled upon this differential equation during the second week of my ODE class in university:
$$ y' - \frac{2}{t+1}y = (t+1)^2$$
And I solved it using the exact procedure in the "Attempt" section below. I checked my work and indeed got the right answer, but as I was solving the problem, I was having difficulty discerning the domain of the function and the equilibrium solutions that arise in the midst of solving such a problem.
I believe there are no equilibrium solutions to take into account in this case, but my main goal of asking this question is for the answerer to outline when to check for equilibrium solutions, how to go about doing it, and providing best practices for constraining both the dependent and independent variables, as well as constants found along the way.
Attempt
Here is my attempt at the solution:
$$ y' - \frac{2}{t+1}y = (t+1)^2 $$
$ t \neq -1 $ is implicitly assumed.
$$ \mu(t) = e^{\int p(t)dt} $$
$$ p(t) = -\frac{2}{t+1} $$
$$ \int p(t)dt = \int -\frac{2}{t+1}dt = -2ln|t+1|+C_1 $$
for $ C_1 \in \mathbb R $.
$$ \mu(t) = e^{-2ln|t+1| + C_1} = e^{C_1}e^{ln|t+1|^{-2}} $$
Let $ C_2 = e^{C_1}$ for $ C_2 \in \mathbb R $, $ C_2 \gt 0 $.
$$ \mu(t) = C_2 e^{ln(\frac{1}{(t+1)^2})} $$
$$ \mu(t) = C_2 \frac{1}{(t+1)^2} $$
Multiplying both sides by the integrating factor $ \mu(t) $,
$$ C_2 \frac{1}{(t+1)^2} y' - C_2 \frac{1}{(t+1)^2} \frac{2}{(t+1)}y = C_2 \frac{1}{(t+1)^2} (t+1)^2 $$
$$ (t + 1)^{-2} y' - 2(t+1)^{-3}y = 1 $$
$$ ((t+1)^{-2}y)' = 1 $$
$$ (t+1)^{-2}y = \int 1 dt $$
$$ (t+1)^{-2}y = t + C_3 $$
for $ C_3 \in \mathbb R $. Dividing both sides by $ \mu(t) $ we get,
$$ \frac{(t+1)^2}{C_2 (t+1)^2} y = \frac{(t+1)^2(t+C_3)}{C_2} $$
$$ y = (t+1)^2(t+C_3) $$
Notes
I'd love the answerer to identify where to check for equilibrium cases in this situation, as well as answer the above bolded questions. I understand equilibrium solutions are to be checked when variables may be separated to solve a differential equation, but I'd love to know all the other cases and have that part be elaborated to me.
Here's the direction field for $ y' = f(t,y) $:

Thanks in advance!
 A: In any non-autonomous system $y' = f(t,y)$, equilibria are found as horizontal asymptotes of the solution $y(t)$. If $t$ represents a "forward-time" variable, then we say $t > t_0$, where $y(t_0)$ is the initial condition, and only look at $\lim\limits_{t\to+\infty} y(t)$.
ex: $y' =-ty \Rightarrow y = y_0e^{-t^2/2}$. Taking $t\to\pm\infty$ gives $y = 0$. Alternatively, $y' = -ty = 0 \Rightarrow t=0$ or $y=0$, and since we assume that $t$ is not always $0$, $y=0$ must be an equilibrium.

In an autonomous system $y' = g(y)$ which doesn't directly depend on $t$, then equilibria are found by solving the equation $y' = 0$, or equivalently $g(y) = 0$, for $y$. This doesn't involve solving the equation, which (among many other features) makes equilibrium analysis much more important for autonomous systems than non-autonomous ones.
ex: $y'= y(1-y) = 0 \Rightarrow y = 0$ or $y = 1$ are equilibria.
A: An equilibrium solution is one in which $y’=0$ is independent of $t$.  In this problem, if you set the derivative equals to $0$ and solve for $y$, any solution will depend on $t$, so there is no equilibrium point.
