Dealing with little o(1) term in ODE I am trying to follow a line in a paper. I will just copy it in here:
(All quantities, i.e. $H,q$ and $\tau$, are functions of $t$.)

As $\tau\to\infty$ we have $q=\frac{1}{2}+o(1)$. Now $dH/dt=-(1+q)H^2$. It follows that $H=\frac{2}{3}t^{-1}(1+o(1))$.

Why does that follow? I can see that truncating the $o(1)$ term, such that $q\approx1/2$ would yield $H\approx\frac{2}{3}t^{-1}$, i.e. the first term of the result.
I however have difficulties to understand why taking the $o(1)$ term in the ODE into account yields precisely this $o(1)$ term in the result after integration. 
 A: It is not the same $o(1)$ term. $o(1)$ represents a function $f$ satisfying $f(\tau)\to 0$ as $\tau\to\infty$, but the function $f$ can change between different instances of the notation $o(1)$. So what the statement means is that $H$ is given as $H(t) = \frac{2}{3}t^{-1} + \frac{2}{3}t^{-1}h(t)$, where $h(t)$ is $o(1)$ as $\tau\to\infty$. Here there is some abuse of notation: we really mean $h(t(\tau))\to 0$ as $\tau\to\infty$; presumably $t$ can also be written as a function of $\tau$. So you'll want to determine this function $h$ (presumably by solving equation, which is separable) and show that it has this asymptotic behavior.
A more detailed answer:
Write $q(t) = \frac{1}{2} + h(t)$, where $h$ is some function satisfying $h(t) = o(1)$ as $\tau\to\infty$. The specific form of $h$ turns out not to matter.
The equation is separable, and solving the initial value problem with initial value specified at $t=0$, we obtain
$$
H(t) = \left(\frac{3t}{2} + \int_0^t h(s)~ds + \frac{1}{H(0)}\right)^{-1}.
$$
We factor out a $\frac{2}{3t}$ from each term, obtaining
$$
H(t) = \frac{2}{3t}\left(1 + \frac{2}{3t}\int_0^t h~ds + \frac{2}{3t}\frac{1}{H(0)}\right)^{-1}.
$$
We wish to show that the right-hand side is $\frac{2}{3t}(1+o(1))$. To see this, we write
$$
\left(1 + \frac{2}{3t}\int_0^t h~ds + \frac{2}{3t}\frac{1}{H(0)}\right)^{-1} = 1 - \frac{\frac{2}{3t}\int_0^t h~ds + \frac{2}{3t}\frac{1}{H(0)}}{1 + \frac{2}{3t}\int_0^t h~ds + \frac{2}{3t}\frac{1}{H(0)}} = 1 + f(t).
$$
We claim that $f(t)$ is $o(1)$ as $t\to\infty$; you haven't provided this detail, but I presume $\tau$ is defined so that $t\to\infty$ if and only if $\tau\to\infty$ in this problem. We simply need to see that $f(t)\to 0$ as $t\to\infty$. To do this we bound each term in the fraction. Since $h(t)\to 0$ as $t\to\infty$, for any $\varepsilon>0$ there exists $T>0$ so that $|h(t)|<\varepsilon$ for $t>T$. Then
$$
\frac{2}{3t}\int_0^t |h(s)|~ds \leq \frac{2}{3t}\int_0^T |h(s)|~ds + \frac{2}{3t}\int_T^t \varepsilon~ds  \leq \frac{2}{3t}\int_0^T |h(s)|~ds + \frac{2\varepsilon}{3}\frac{t-T}{t}.
$$
In the right-most expression, the second term is bounded by $C\varepsilon$ as $t\to\infty$, where $C$ is some constant independent of $t$. The first term is bounded by $\frac{1}{t}$ times a constant (in $t$). So for any $\varepsilon$, letting $t\to\infty$ we find that 
$$
\lim_{t\to\infty}\frac{2}{3t}\int_0^t |h(s)|~ds \leq C\varepsilon.
$$
Since $\varepsilon$ was arbitrary it follows that
$$
\lim_{t\to\infty}\frac{2}{3t}\int_0^t h(s)~ds = 0.
$$
As for $\frac{1}{t}\frac{1}{H(0)}$, obviously this tends to $0$ as $t\to\infty$. Therefore taking limits, we find that
$$
\lim_{t\to\infty} f(t) =  -\frac{\lim_{t\to\infty}(\frac{2}{3t}\int_0^t h~ds + \frac{2}{3t}\frac{1}{H(0)})}{\lim_{t\to\infty}(1 + \frac{2}{3t}\int_0^t h~ds + \frac{2}{3t}\frac{1}{H(0)})} = \frac{0}{1} = 0.
$$
Therefore $f(t) = o(1)$ as $t\to\infty$.
