Behave of the solution as $t\to\infty$ Consider the following ODE $$y'+7y=t+e^{-2t}$$ It is easy that $y=\frac{7t-1}{49}+\frac{1}{5e^{2t}}+\frac{c}{e^{7t}}$, where $c$ is a constant. I want to determine how solution behaves as $t\to\infty.$ I thought I just need to take the limit and this showed that $y\to\infty.$ Since the answer in the book is $y=\frac{7t-1}{49}$ as $t\to\infty.$ Should I graph the solution to get this answer, I meant should I graph the direction field to answer this question? Any advice?
 A: Just for fun, let's throw some unnecessary high-powered machinery at this problem and use a method inspired by (or perhaps even equal to) the method of dominant balance.
Looking at the right-hand side of the equation, we can see that at $t\to\infty$, the $t$ term dominates. For this reason, let's suppose that
$$
7y(t) \sim t\Longrightarrow y\sim\frac{t}{7}.
$$
Plugging in $y=t/7$ to both sides yields
$$
\frac{1}{7}+7\frac{t}{7}\sim t+e^{-2t}.
$$
It is certainly true that the
$$
\lim_{t\to\infty}\frac{\mbox{left-hand side}}{\mbox{right-hand side}}=1,
$$
so it seems like this was a good choice to make.
Next, we suppose that
$$
y=\frac{t}{7}+\epsilon(t),
$$
where $\epsilon(t)\ll t$, which is to say,
$$
\lim_{t\to\infty}\frac{\epsilon(t)}{t}=0.
$$
Plugging this into the differential equation yields
$$
49\epsilon+7\epsilon'=-1+7e^{-2t}.
$$
On the right-hand side, the dominant term is the $-1$, so let's suppose that
$$
\epsilon\sim -1/49.
$$
Pugging this in, we get
$$
-1+0\sim-1+7e^{-2t},
$$
and the sides still clearly balance.
Thus, we arrive at an asymptotic solution of the form
$$
y(t) \sim \frac{t}{7}-\frac{1}{49}.
$$
We could stop there, because the rest of the terms are likely to be exponentially small, but let's see how much of this expansion we can get!

So, we proceed again by guessing that
$$
\epsilon(t) = -\frac{1}{49}+\epsilon_2(t),
$$
where again, $\epsilon(t)\ll\epsilon_2(t)$ as $t\to\infty$.  Plugging this into the differential equation for $\epsilon$, we get
$$
7\epsilon_2+\epsilon_2'=e^{-2t}.
$$
Now there is only one term on the right, so we guess that
$$
\epsilon_2\sim ae^{-2t}
$$
Plugging this in, we get
$$
5ae^{-2t}\sim e^{-2t},
$$
which again balances!  If, in the next step, we actually want these terms to cancel, then we should choose $a=1/5$.
So, guess the next iteration to be
$$
\epsilon_2(t) = \frac{1}{5}e^{-2t} +\epsilon_3(t),
$$
again where $\epsilon_2\ll\epsilon_3$.  Plugging this in, we arrive at
$$
\epsilon_3'=-7\epsilon_3.
$$
We've finally gotten to the point where we know the solution, i.e.,
$$
\epsilon_3(t) = ce^{-7t}.
$$
Putting everything back together, we arrive at the asymptotic expansion
$$
y(t) \sim \frac{t}{7}-\frac{1}{49}+\frac{1}{5}e^{-2t}+ce^{-7t}.
$$
Of course, it turns out that this is the exact solution, but what's the fun in figuring out exact solutions when you can perform an asymptotic analysis!
