Behavior of a non-linear differential equation Let us consider the following differential equation
$$
\dot{x}(t)=a - b\sin(x(t)), \quad a,b\in\mathbb{R}.
$$

My question. Suppose $a>|b|$ and $x(0)=x_0\in\mathbb{R}$. Can the solution to the above equation be written in the form
  $$
x(t) = at + r(a,t),
$$
  where the term $r(a,t)$ is such that $r(a,t)\to 0$ for $a\to \infty$?

PS: Of course, the explicit solution can be computed via symbolic software tools, such as Wolfram Alpha. However, the symbolic expression (see here) looks quite messy and does not give much information on the behavior of the solution for $a\to\infty$...
 A: Consider the case $a > |b|$, so the solutions are unbounded.  The equation is separable, and we get implicit solutions of the form
$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)}  $$
which we can expand in a series in $1/a$ (uniformly convergent in $s$).  Absolute convergence justifies interchanging sum and integral, so (for fixed $x$ and $b$)
$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (b \sin(s))^k a^{-1-k}\cr
&= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}b^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$
Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.
A: Take the ODE
$$
\tag{$*$}
\dot{x}(t) = a - b \sin(x(t))
$$
on the one-dimensional torus $\mathbb{R}/2 \pi \mathbb{Z}$.  The vector field has no zeros, so there is a unique periodic orbit, with period
$$
\int\limits_{0}^{2 \pi} \frac{d\xi}{a - b \sin(\xi)} =
\frac{2 \pi}{\sqrt{a^2 - b^2}}.
$$
Return now to $(*)$ on the real line $\mathbb{R}$.  We have
$$
x\!\left(t + \frac{2 \pi}{\sqrt{a^2 - b^2}}\right) = x(t) + 2 \pi \quad \text{for all }t \in \mathbb{R}.
$$
Put
$$
h(t) := x(t) - \sqrt{a^2 - b^2} t.
$$
$h$ is easily seen to be periodic, with period $2\pi/\sqrt{a^2 - b^2}$.  Consequently, we have, for any solution $x(\cdot)$ to $(*)$,
$$
x(t) = \sqrt{a^2 - b^2} t + h(t).
$$
So,
$$
r(a, t) = (\sqrt{a^2 - b^2} - a)t + h(t),
$$
which, for a fixed $t$, converges to $h(t)$ as $a \to \infty$.
