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 x_\varepsilon(0)\in\mathbb{R}, \end{align} where $\varepsilon$ is a real constant. Let us define $$ \Delta(t,\varepsilon):= |\cos({x}_{\varepsilon}(t))-\cos({x}_{0}(t))| $$ Clearly, note that $\Delta(t,0)\equiv 0$.
Is it true that $\Delta(t,\varepsilon)$ is 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 the answer is no, is it possible to find upper bound such that $\Delta(t,\varepsilon)\le K |\varepsilon| p(t)$ where $p(t)$ is a polynomial function of $t$?]
Numerical simulations seem to confirm this claim. Any help towards a theoretical confirmation (or rejection) of this claim is more than welcome! Many thanks!
An initial (incomplete) attempt. We can rewrite $\Delta(t,\varepsilon)$ as $$ \Delta(t,\varepsilon)= 2\left|\sin\left(\frac{{x}_{\varepsilon}(t)-{x}_{0}(t)}{2}\right)\sin\left(\frac{{x}_{\varepsilon}(t)+{x}_{0}(t)}{2}\right)\right|, $$ so that $$ \Delta(t,\varepsilon)\le 2\left|\sin\left(\frac{{x}_{\varepsilon}(t)-{x}_{0}(t)}{2}\right)\right|\le |{x}_{\varepsilon}(t)-{x}_{0}(t)|. $$ So now the problem is perhaps a bit simplified (?) and boils down to find an upper bound to $|{x}_{\varepsilon}(t)-{x}_{0}(t)|$. However, I don't know how to proceed from here.