Electro-Mechanics with Source

$\textcolor{#20A900}{Fun Problem}$
As always, it is a pleasure to solve your problems. Thank you for another good one.

The basic equations are $\textcolor{blueviolet}{BvD-10sin(5t) -iR-L \frac{di}{dt}=0}$ $\textcolor{blueviolet}{F-BiD=m \frac{dv}{dt}}$ After substituting value $10v-10sin(5t) -2i-\dot{i}=0$ $\dot{v}+10i=2$ Solving both equations, gives the differential equation in terms of $v$ $\textcolor{midnightblue}{10v+0.2\dot{v}+0.1\ddot{v}=10sin(5t) +0.4}$ Solving this differential equation and using initial conditions $\textcolor{#624F41}{v(0) =0}$ $\textcolor{#624F41}{\dot{v}(0)=2}$ to find the values of arbitrary constant in the solution of differential equation. And after substituting the value of arbitary constants gives the solution of differential equation as $\Large \textcolor{forestgreen}{v=1.31004sin(5t) -0. 439757303e^{-t}sin(9.94987t) +0.174672e^{-t}cos(9.94987t) -0.174672cos(5t) +0.04}$ Substituting $\large v=\frac{dx}{dt}$ $\Large \textcolor{#BA33D6}{\frac{dx}{dt}=1.31004sin(5t) -0. 439757303e^{-t}sin(9.94987t) +0.174672e^{-t}cos(9.94987t) -0.174672cos(5t) +0.04}$ $\Large \int_{x(0) }^{x(5)} dx=\int_{t=0}^{t=5}(1.31004sin(5t) -0. 439757303e^{-t}sin(9.94987t) +0.174672e^{-t}cos(9.94987t) -0.174672cos(5t) +0.04)dt$ $\large \textcolor{#3D99F6}{\boxed{x(5) =0.165087}}$