Solution:

The given equation $\frac{d i}{d t}+\frac{R i}{L}=\frac{E}{L}$ is linear of the type $\frac{d y}{d x}+P y=Q$

$\therefore$ Its solution is $i e^{\int R / L \ d t}=\int e^{\int R / L \ d t} \cdot \frac{E}{L} \cdot d t+c$

$\begin{aligned} i . e^{R t / L} &=\frac{E}{L} \int e^{R t / L} d t+c=\frac{E}{L} \cdot e^{R t / L} \frac{L}{R}+c \\ &=\frac{E}{R} e^{R t / L}+c \end{aligned}$

When $t=0$ and $i=0 \therefore c=-\frac{E}{R}$

$\therefore$ i. $e^{R t / L}=\frac{E}{R} e^{R t / L}-\frac{E}{R}$

$\therefore i=\frac{E}{R}\left(e^{R t / L}-1\right)$

$\therefore i=\frac{E}{R}\left(1-e^{-R t / L}\right)$