Answer
(a) $i=(\dfrac{V}{R})+ce^{-Rt/L}$;
(b) $i=(\dfrac{V}{R})-(\dfrac{V}{R})e^{-Rt/L}$;
(c) $i=\dfrac{V}{R}$ is a solution of the differential equation
$\dfrac{di}{dt}+\dfrac{R}{L}i=\dfrac{V}{L}$
$i=ce^{-Rt/L}$ is a solution equation $\dfrac{di}{dt}+\dfrac{R}{L}i=0$
Work Step by Step
(a) $\dfrac{di}{dt}+\dfrac{R}{L}i=\dfrac{V}{L}$
The integrating factor is: $I=e^{\int (R/L) dt}=e^{Rt/L}$
Now, $e^{Rt/L}[\dfrac{di}{dt}+\dfrac{R}{L}i]=e^{Rt/L}(\dfrac{V}{L})$
This gives: $\int [e^{Rt/L}i]'=\int (\dfrac{V}{R}) e^{Rt/L} dt $
$\implies i=(\dfrac{V}{R})+ce^{-Rt/L}$
(b) Apply the initial conditions.
$ i=(\dfrac{V}{R})+ce^{-(Rt/L)} \implies c=-\dfrac{V}{R}$
or, $i=(\dfrac{V}{R})+ce^{-Rt/L} \implies i=(\dfrac{V}{R})-(\dfrac{V}{R})e^{-Rt/L}$
(c) consider $\dfrac{di}{dt}+\dfrac{R}{L}i=\dfrac{V}{L}$
or, $\dfrac{di}{dt}+\dfrac{R}{L}(\dfrac{V}{R})=\dfrac{V}{L}$
or, $\dfrac{di}{dt}=0$
we need to prove that $i=ce^{-Rt/L}$ is a solution equation $\dfrac{di}{dt}+\dfrac{R}{L}i=0$
or, $\dfrac{di}{dt}=c(-\dfrac{R}{L}) e^{-(Rt/L)}$
Thus, we have $\dfrac{di}{dt}+(\dfrac{R}{L})i=0$
