Perturbed behavior of a differential equation Let $a$, $b$ be two real positive parameters with $a>b$, and consider the following nonlinear differential equation:
\begin{align}
\dot{x}_{\varepsilon}(t) = a - b\sin(x_{\varepsilon}(t))+\varepsilon,\quad t\ge 0,\ x_\varepsilon(0)\in\mathbb{R},
\end{align}
where $\varepsilon$ is a positive real constant.
Let us define 
$$
\Delta(t,\varepsilon):= |{x}_{\varepsilon}(t)-{x}_{0}(t)|,
$$
and note that $\Delta(t,0)\equiv 0$.

My question. Suppose that $x_\varepsilon(0)=x_0(0)$. Is $\Delta(t,\varepsilon)$ linearly bounded in $t\ge 0$ and $\varepsilon$, that is
  $$
 \Delta(t,\varepsilon)\le K \varepsilon t,
$$
  where $K$ being a suitable constant? [In case of positive answer, if $x_\varepsilon(0)\ne x_0(0)$, is it possible to find an upper bound (similar to the above one) which takes into account the initial conditions?]

 A: $\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\renewcommand{\th}{\theta}
\newcommand{\om}{\omega}
\newcommand{\R}{\mathbb{R}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ 
Let $x=x(a,t)$ be a solution to the diff. eq.
\begin{equation*}
 \dot{x}(a,t) = a - b\sin x(a,t). \tag{0}
\end{equation*}
The variables $t$ and $x$ are separable. Indeed, this diff. eq. can be rewritten as 
\begin{equation*}
 \int\frac{dx}{a - b\sin x}=\int dt, 
\end{equation*}
whence 
\begin{equation*}
 \frac2\om\, \tan ^{-1}\frac{b-a \tan(\frac{x}{2})}{\om}=t+c
\end{equation*}
and 
\begin{equation*}
 x(a,t)=f(a,t+c), \tag{1}
\end{equation*}
for some real constant $c$, where 
\begin{equation*}
 f(a,t):=2 \tan ^{-1}\frac{b+\om \tan (\frac{1}{2}\om t)}{a}
\end{equation*}
and 
\begin{equation*}
 \om:=\om(a):=\sqrt{a^2-b^2}. 
\end{equation*}
Letting 
\begin{equation*}
 T:=2\pi/\om, \tag{2}
\end{equation*}
note that $f(a,t)$ is smooth in $t\in(-T/2,T/2)$, with the partial derivative
\begin{equation*}
 f'_t(a,t)=\frac{a\om^2}{a^2-b(b \cos \om t+\om \sin \om t)}\in\Big(\frac{\om^2}{a+b},\frac{\om^2}{a-b}\Big) \tag{3}
\end{equation*}
in $t$. 
Also, $f(a,t)\to\pm\pi$ as $\pm t\uparrow T/2$. 
Thus (and this is the crucial point), the function $(-T/2,T/2)\ni t\mapsto f(a,t)$ can be extended to a smooth function $\R\ni t\mapsto f(a,t)$ such that 
\begin{equation*}
 \psi(a,t):=f(a,t)-\om(a) t
\end{equation*}
is periodic in $t$ with period $T$ as in (2), and then (1) will define the general solution $x$ to the diff. eq. (0). 
The graph of $f$ for $a=3,b=2.5$ is shown here: 

Fix any real $x_0$. In view of (3), there is a unique real $c(a)$ such that 
\begin{equation*}
 f(a,c(a))=x_0, \tag{4} 
\end{equation*}
and the the formula 
\begin{equation*}
 x(a,t)=f(a,t+c(a))=\om(a)t+\psi(a,t+c(a))
\end{equation*}
will define the unique solution to (0) satisfying the initial condition $x(a,0)=x_0$. 
For each real $t$, 
\begin{equation*}
 x'_a(a,t)=\frac d{da}\,f(a,t+c(a))=\om'(a)t+\psi'_a(a,t+c(a))+\psi'_t(a,t+c(a))c'(a). 
\end{equation*}
By (4), the implicit function theorem, and (3),
\begin{equation*}
 |c'(a)|=|f'_a(a,c(a))/f'_t(a,c(a))|\ll|f'_a(a,c(a))|\ll\om'(a)|c(a)|+|\psi'_a(a,c(a))|;  
\end{equation*}
the constants in $\ll$ may depend on $a$ and $b$, but not on $t$. 
Since $\psi(a,t)$ is smooth in $(a,t)$ and periodic in $t$, it follows from the last two displays that 
\begin{equation*}
 |x'_a(a,t)|\ll t+1. \tag{5} 
\end{equation*}
So, for $t>1$ we have $|x'_a(a,t)|\ll t$ and hence 
\begin{equation*}
 |x(a+\ep,t)-x(a,t)|\ll\ep t. \tag{6}
\end{equation*}
It remains to show that (6) holds for $t\in[0,1]$ as well. This is now easy:  because of the smoothness of $f(a,t)$ in $(a,t)$ and the above bound on $|c'(a)|$, for $t\in[0,1]$ 
\begin{equation*}
 |x(a+\ep,t)-x(a,t)|\le\int_0^t|f'_t(a+\ep,s+c(a+\ep))-f'_t(a,s+c(a))|\,ds\ll \int_0^t\ep\,ds=\ep t. 
\end{equation*}
So, (6) holds for all $t\ge0$. 
A: First write the differential equation as an integral equation $\newcommand{\ve}{{\varepsilon}}$
$$
x_\ve(t)=x_0+(a+\ve)t-b\int_0^t \sin x_\ve(s) ds. $$
We deduce
$$ x_\ve(t)-x_0(t)= \ve t-b\int_0^t\big(\; \sin x_\ve (s)-\sin x_0(s)\;\big) ds $$
so (for $t\geq 0$)
$$ \Delta(t,\ve)\leq \ve t +b\int_0^t \Delta(s,\ve) ds. $$
Gronwall's inequality now implies
$$\Delta(t,\ve)\leq\ve t +b\ve e^t\int_0^t  s e^{-s} ds = \ve t +b\ve e^t \Big(\; 1-(t+1)e^{-t}\;\Big). $$
