First write the differential equation as an integral equation $\newcommand{\ve}{{\varepsilon}}$

$$ x_\ve(t)=x_0+(a+\ve)t-b\int_0^t x_\ve(s) ds. $$

We deduce

$$ x_\ve(t)-x_0(t)= \ve t-b\int_0^t\big(\; x_\ve (s)-x_0(s)\;\big) ds $$

so (for $t\geq 0$)

$$ \Delta(t,\ve)\leq \ve t +\int_0^t \Delta(s,\ve) ds. $$

Gronwall's inequality now implies

$$\Delta(t,\ve)\leq\ve t +b\ve e^t\int_0^t s e^{-s} ds = \ve t +b\ve e^t \Big(\; 1-(t+1)e^{-t}\;\Big). $$

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

$$\Delta(t,\ve)\leq\ve t +b\ve e^t\int_0^t s e^{-s} ds = \ve t +b\ve e^t \Big(\; 1-(t+1)e^{-t}\;\Big). $$