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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