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](http://www.wolframalpha.com/input/?i=dx(t)%2Fdt%20%3D%20a%20-%20b*sin(x))) looks quite messy and does not give much information on the behavior of the solution for $a\to\infty$...



Behavior of a non-linear differential equation
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.


