Answer
See answers below
Work Step by Step
Using RC Circuit we get:
$$\frac{di}{dt}+\frac{R}{L}=\frac{R}{L}$$
The integrating factor is:
$$I=e^{\int \frac{R}{L}dt}=e^{\frac{R}{L}t}$$
Multiplying by integrating factor:
$$e^{\frac{R}{L}t}\frac{di}{dt}+e^{\frac{R}{L}t}\frac{R}{L}=e^{\frac{R}{L}t}\frac{R}{L}$$
$$e^{\frac{R}{L}t}\frac{di}{dt}+e^{\frac{R}{L}t}\frac{R}{L}=e^{\frac{R}{L}t}E_0\sin wt$$
$$\frac{d}{dt}(e^{\frac{R}{L}q})=e^{\frac{R}{L}t}\frac{E_0}{L}\sin wt$$
Integrating both sides:
$$e^{\frac{R}{L}q}=c-\frac{E_0e^{\frac{R}{L}t}(Lw \cos (wt)-R \sin (wt))}{L^2w^2+R^2}$$
$$q=ce^{-\frac{R}{L}t-\frac{E_0}(Lw \cos (wt)-R \sin (wt))}{L^2w^2+R^2}$$
As $t \rightarrow \infty$, the transient part of the solution is $ce^{\frac{R}{L}t}$
and the steady-state solution is: $-\frac{E_0(Lw \cos (wt)-R \sin (wt))}{L^2w^2+R^2}$
