Calculus

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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 -

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 -

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

dP/P=-2 dt
lnP=-2t
P=e^-2t

• Calculus -

thank you!!

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