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 to the equation $\dfrac{di}{dt}+\dfrac{R}{L}i=0$
Work Step by Step
a) Here, we have
$\dfrac{di}{dt}+\dfrac{R}{L}i=\dfrac{V}{L}$
The integrating factor is: $e^{\int (R/L) dt}=e^{Rt/L}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [e^{Rt/L}i]'=\int (\dfrac{V}{R}) e^{Rt/L} dt $
or, $i=(\dfrac{V}{R})+ce^{-Rt/L}$ ...(2)
b) Applying the initial conditions in the above equation, we get
$ c=-\dfrac{V}{R}$
Thus, equation (2) becomes: $i=(\dfrac{V}{R})-(\dfrac{V}{R})e^{-Rt/L}$
c) Here, we have
$\dfrac{di}{dt}+\dfrac{R}{L}i=\dfrac{V}{L} \implies \dfrac{di}{dt}+\dfrac{R}{L}(V/R)=\dfrac{V}{L}$
or, $\dfrac{di}{dt}=0$
We need to show that $i=ce^{-Rt/L}$ is a solution to the equation $\dfrac{di}{dt}+\dfrac{R}{L}i=0$
After taking the derivative we get
$\dfrac{di}{dt}=c(-\dfrac{R}{L}) e^{-Rt/L}$
Thus, we have $\dfrac{di}{dt}+\dfrac{R}{L}i=0$
