#### Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 10

Answer: $i=\frac{E}{R}\left[1-e^{\left(-\frac{K}{L}\right) t}\right]$

Given:  $L \frac{d i}{d t}+R i=E$

To find: To prove that $i=\frac{E}{R}\left[1-e^{\left(-\frac{R}{L}\right) t}\right]$

Hint: Integrate the linear differential equation to find integral factor.

Solution: $L \frac{d i}{d t}+R i=E$

Dividing by L

$=\frac{d t}{d t}+\frac{R}{L} i=\frac{E}{L}$

This is a linear differential equation and comparing it with

$=\frac{d y}{d x}+P y=Q$

Such that,

$=P=\frac{R}{L^{\prime}}, Q=\frac{E}{L}$

We know that

\begin{aligned} &\text { I.F. }=e^{\int p d t} \\\\ &=e^{\int \frac{R}{L} d t} \\\\ &=e^{\left(\frac{R}{L}\right) t} \end{aligned}

So, the solution is given by

\begin{aligned} &i(I f)=\int Q(I f) d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\int \frac{E}{L}\left[e^{\left(\frac{R}{L}\right) t}\right] d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{L} \times \frac{L}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \end{aligned}

\begin{aligned} &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \\\\ &=>i=\frac{\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C}{\left[e^{\left.\left(\frac{R}{L}\right) t\right]}\right.} \end{aligned}

\begin{aligned} &\text { => } i=\frac{E}{R}+\frac{C}{\left[e^{\left(\frac{R}{L}\right) t}\right]} \\\\ &=>i=\frac{E}{R}+C\left[e^{-\left(\frac{R}{L}\right) t}\right] \ldots(i) \end{aligned}

Initially no current passes through the circuit so,i=0, t=0

\begin{aligned} &=>0=\frac{E}{R}+C e^{0} \\\\ &=>0=\frac{E}{R}+C \\\\ &=>C=-\frac{E}{R} \end{aligned}

Substituting values of C in equation (i)

\begin{aligned} &=>i=\frac{E}{R}-\frac{E}{R}\left[e^{-\left(\frac{R}{L}\right) t}\right] \\\\ &=>i=\frac{E}{R}\left[1-e^{-\left(\frac{R}{L}\right) t}\right] \end{aligned}