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 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?]

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?]