Answer
$i=\frac{E_0}{R}(1-e^{\frac{R}{L}t})$
Work Step by Step
Using RL Circuit, we get:
$$\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}$$
The integrating factor is:
$$I=e^{\frac{R}{L}dt}=e^{\frac{R}{L}t}$$
Multiplying by integrating factor:
$$e^{\frac{R}{L}t}\frac{di}{dt}+e^{\frac{R}{L}t}\frac{R}{L}i=e^{\frac{R}{L}t}\frac{E}{L}$$
$$\frac{d}{dt}(e^{\frac{R}{L}t}q)=e^{\frac{R}{L}t}\frac{E}{L}$$
Integrating both sides:
$$e^{\frac{R}{L}t}q=c+e^{\frac{R}{L}t}\frac{E_0}{R}$$
$$q=ce^{-\frac{R}{L}t}+\frac{E_0}{R}$$
Since $i(0)=0$, we have:
$$0=c+\frac{E_0}{R}$$
sOLVE FOR $C$:
$$ \rightarrow c=-\frac{E_0}{R}$$
Thus, the current flowing in an RL circuit is:
$$i=\frac{E_0}{R}(1-e^{\frac{R}{L}t})$$
