I'm having trouble with this, here's my attempt:

$\displaystyle V=V_{R}+V_{L}$

$\displaystyle = IR + L \frac{dI(t)}{dt}$

Now introduce f(t) for the integratino factor, divide by L to get the first order on its own

$\displaystyle f(t)\frac{V}{L} = f(t)\frac{IR}{L} + f(t)\frac{dI(t)}{dt}$

$\displaystyle f '(t) = \frac{Rf(t)}{L}$

$\displaystyle f(t) = e^{\int \frac{R}{L} dt}$

So far this seems somewhat logical, as the final answer has the minus factor in it.

Continuing:

$\displaystyle e^{\frac{Rt}{L}}\frac{V}{L} = \frac{d}{dt} [e^{\frac{Rt}{L}} I]$

Integrating this gets me back to

$\displaystyle V = IR$

as theres nothing i can do with the exponentials

The answer i should be getting is:

$\displaystyle I = I_{0} [1-e^{-\frac{Rt}{L}}] $

Hope you can help me out