some questions about differential equations when solving 1st order linear differential equations, we have some common method. 
I have some questions about the common methods.
for example 
$\dfrac{dy}{-y + 5} = dt$ , why can we integrate on both side to get the equation of $y$? More specifically, we have different things to integrate, $y$ and $t$ respectively, why can we just integrate on both side? What's the underlying reason here?
Also when we get $\,\ln |5 - y| =  -t + C,\,$ why can we immediately get $\,5- y = e^{-t + C}\,$ and get rid of the absolute sign? There is no information about the range of $y$, why can we do that? Thanks
 A: Your confusion stems from a lack of knowledge of what the notation means. When we write
$$
\frac{dy}{-y+5} = dt
$$
what we mean is 
$$
\frac{dy}{dt} \frac{1}{-y+5} = 1
$$
and we then integrate both sides in $t$, (recall $y=y(t)$). The reason it looks like we are integrating the left side in terms of $y$ is because of the chain rule.
The equation being separable means we can write it like
$$
f(y)dy = g(t)dt
$$
which means
$$
\frac{dy}{dt} f(y(t)) = g(t).
$$
But by the chain rule , if $F,G$  antiderivatives of $f,g$, then we have
$$
\frac{d}{dt}(F(y(t)) = f(y(t))\frac{dy}{dt}
$$
and integrating in $t$ gives
$$
F(y(t)) = G(t)+C
$$
which looks like you integrated in $y$ on the left and $t$ on the right in the sloppy notation.
As for why you can get rid of absolute values, you can't. You must consider $5-y = e^{-t+C}$ as well as $y-5 = e^{-t+C}$. This gives us
$$
y = 5-e^C e^{-t},~~ y= 5 + e^C e^{-t}
$$
as valid solutions. But note that $e^C$ is an arbitrary positive constant, and $-e^C$ is an arbitrary negative constant, so both solutions may be conglomerated into the shorter form
$$
y = 5 + ce^{-t}
$$
where $c$ is an arbitrary nonzero constant.
If the original question was given as $y' = -y+5$, that is YOU divided by $-y+5$, then you must also check that the constant solution $y=5$ works, so that $y=5+ce^{-t}$ is valid for all $c \in \mathbb{R}$. If the equation was given in the form with $-y+5$ in the denominator already, i.e. you did not introduce division by $-y+5$, then we leave the answer with $c \neq 0$ as the statement would not make sense if $y=5$.
A: Actually, here how it goes, you have:
$$
\frac{1}{5-y}\frac{dy}{dt}=1,
$$
and you integrate both sides with respect to $t$ and then use the substitution in the left hand side:
$$
\int \frac{1}{5-y(t)}\frac{dy}{dt}dt=\int \frac{1}{5-y}{dy}=\int dt,
$$
from which
$$
-\ln |5-y|=t+C.
$$
After you get
$$
|5-y|=e^{-t-C}=C_1e^{-t},
$$
where $C_1$ is supposed to be positive.
After that
$$
5-y=\pm C_1e^{-t}=C_2e^{-t},
$$
where $C_2$ can be now arbitrary (except for zero, however you can check that $y=5$ is a solution of $y'=5-y$, which means that $C_2$ is an absolutely arbitrary constant).
