Posted by **Daniel** on Thursday, December 4, 2008 at 6:45pm.

Find the solution subject to the initial conditions.

dP/dt = -2P P(0)=1

dP/-2P = dt

then do i take the antiderivative? what would i do next??

- Calculus -
**drwls**, Thursday, December 4, 2008 at 6:51pm
Yes.

ln P = -2 t + C

P = e^(-2t + C)

= C' e^-2t where C' is a different arbitrary constant

C' = 1 since P(0) = 1

Therefore P = e^-2t

- Calculus -
**bobpursley**, Thursday, December 4, 2008 at 6:53pm
What if P= Ce^at + D

dP/dt= aCe^at and if D is zero..

dP/dt= ac e^at=aP

now if a is -2

dp/dt=-2P Hmmm.

P(O)=1=Ce^a0=C so C=1

Now for your question...

dP/P=-2 dt

lnP=-2t

P=e^-2t

- Calculus -
**Daniel**, Thursday, December 4, 2008 at 6:59pm
thank you!!

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