How to solve this differential equation corresponding to the RL circuit?
$$L\frac{di_d}{dt} = \sqrt{2} U_2 \sin(\omega t) - Ri_d$$
$$\frac{di_d}{dt} + \frac{R}{L}i_d = \frac{\sqrt{2} U_2}{L} \sin(\omega t)$$
Multiplying both sides by e^(Rt/L),
$$e^{\frac{Rt}{L}}\frac{di_d}{dt} + e^{\frac{Rt}{L}}\frac{R}{L}i_d = e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t)$$
By product rule of differentiation,
 $$\frac{d(e^{\frac{Rt}{L}}i_d)}{dt} = e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t)$$
$$e^{\frac{Rt}{L}}i_d = \int e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t) dt$$
$$i_d = e^{\frac{-Rt}{L}}\int e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t) dt$$

Integrating and substituting your initial conditions, you should get your solution for i.