Problem solved.

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- Jun 28th 2006, 04:03 PMDaeIntegration Confusion
Problem solved.

- Jun 28th 2006, 04:20 PMJameson
I don't follow your work, but using the integrating method goes like this.

$\displaystyle \frac{dM}{dt}=-kM$

$\displaystyle \frac{1}{M}dM=-kdt$

$\displaystyle \ln|M|=-kt+C$

$\displaystyle M=Ce^{-kt}$ - Jun 28th 2006, 04:44 PMDae
Thanks Jameson, I think I get that.

- Jun 28th 2006, 06:07 PMThePerfectHacker
Dae, do not EVER delete your original post once your question was answered.

-=USER WARNED=- :mad: