Current using kirchhoff's loop rule

[8614smith]

Homework Statement

By applying kirchhoffs rules, show that the in the RL circuit is given by: I=\frac{\epsilon}{R}\left(1-e^{-\frac{Rt}{L}}\right)

Homework Equations

kirchhoff's second rule: \epsilon-IR-L\frac{dI}{dt}=0

The Attempt at a Solution

\epsilon=IR+L\frac{dI}{dt}
{\epsilon}dt=IRdt+LdI
\frac{\epsilon}{I}dt-Rdt=L\frac{1}{I}dI
then i thought to integrate both sides, LHS w.r.t t and RHS w.r.t I, but doing this on paper doesn't get to the right answer, am i going wrong somewhere with the rearranging??

 

[vela]

You can't integrate it the way you have it written. You want to get all the I's on one side and all the t's on the other first, then you can integrate it.

You could also just plug the solution in and show it satisfies the differential equation instead of solving the equation.