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Given the differential equation : $$E-L\frac{di}{dt}=Ri$$ where $i$ is a function of $t$ and $E$, $L$ and $R$ are constants. If at time $t = 0$, the current, $i$, is zero

i need to use the integrating factor method to show that at time $t$,$$i=\frac{E}{R}(1-e^{-\frac{R}{L}t})$$

Can i get some help on how to start solving this? Thanks.

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$$RI+L\frac{dI}{dt}=E\tag{1}$$ $$\frac{dI}{dt}+\frac{R}{L}I=\frac{E}{L}$$ by finding integration factor $\displaystyle e^{\int\frac{R}{L}dt}=e^{\frac{R}{L}t}$ the general solution is $$Ie^{\frac{R}{L}t}=\int e^{\frac{R}{L}t}\frac{E}{L}dt+C$$ or $$I=\frac{E}{R}+Ce^{-\frac{R}{L}t}$$ in $t=0$ the current is zero $I(0)=0$ so $\displaystyle C=-\frac{E}{R}$ then particular sollution will be $$I=\frac{E}{R}(1-e^{-\frac{R}{L}t})$$

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    How did you get this solution: $$I=\frac{E}{R}+Ce^{-\frac{R}{L}t}$$ – 2017-02-07
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    After integration, Mutiple sides by $e^{-\frac{R}{L}t}$ – 2017-02-07
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    Im find it hard to integrate that $\int \:e^{\frac{R}{L}t}\cdot \frac{E}{L}\:dt $.Which method did you use? – 2017-02-07
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    Not Hard. $\int e^{\alpha t}dt=\frac{1}{{\alpha}}e^{\alpha t}$ and $\frac{E}{L}$ is a number which you bring it out of integral. – 2017-02-07