# Laplace transform problem RL circuit

I'm having trouble solving an RL circuit using the Laplace Transform.

There's just a 2H inductor in series with a 5M ohm resistor. The inductor is initially charged to 1.25A.

So... Here's what I've done: $$v_L(t) = v_R(t)\\ 2i_L'(t) = 5Mi_L(t)$$ Now taking the Laplace Transform, $$2(I_L(s)s - 1.25) = 5M I_L(s)\\ 2.5 = 2I_L(s)s - 5M I_L(s)\\ 2.5 = I_L(s)(2s - 5M)\\ I_L(s) = \frac{2.5}{2s-5M}\\ I_L(s) = \frac{1.25}{s-2.5M}$$ Now taking the inverse Laplace transform, $$i_L(t) = 1.25e^{2.5Mt}$$ This answer is wrong though, and there should be negative $$i_L(t) = 1.25e^{-2.5Mt}$$

For the life of me, I can't work out where that negative comes from in the maths.

I'd really appreciate some help.

Thanks!

Oh, I should add that the $M$ is Mega (so $10^6$).

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For the system you wrote at the beginning your solution is correct. – Artem Jun 7 '12 at 5:15

Either take $v_R = v_L$, in which case $i_R = -i_L$. This would be the usual convention (ie, current 'enters' the device through the $+$ terminal).
Or you could take $i_R = i_L$, in which case you will need to take $v_R = -v_L$.