I am looking at the differential equation
$x'(t)=x(t) (1+s(t))$.
and I want to make a statement about the behaviour of its solution $x(t)$ for large $t$, knowing that for large $t$
$s(t)=\frac{1}{2}-\frac{1}{2t}+\text{terms decaying like }t^{-2}\text{ or faster}.$
Naively I would simply drop the higher order terms, since I know that for large enough $t$ those should be negligible. Setting $s(t)\approx\frac{1}{2}-\frac{1}{2t}$ and solve the resulting differential equation, yields
$x(t)\propto e^{3t/2}/\sqrt t$.
Is this way of approximating the differential equation and its solution for large $t$ legal? If not, why not, and what else could I say about the solution $x(t)$ for large $t$, given the asymptotic behaviour of $s(t)$?
If $x > 0$, you can write your differential equation as
$$ \dfrac{d}{dt} \log(x(t)) = 1 + s(t)$$ so $$ \log(x(t)) = c + \int (1+s(t))\; dt = c + \frac{3}{2} t - \frac{1}{2} \log t + O(t^{-1})$$ That is, there exist $T$ and $\epsilon$ such that for all $t > T$,
$$ c + \frac{3}{2} t - \frac{1}{2}\log t - \frac{\epsilon}{t} < \log(x(t)) < c + \frac{3}{2} t - \frac{1}{2}\log t + \frac{\epsilon}{t}$$
Now take the exponential...