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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

    Now for your question...
    dP/P=-2 dt
    lnP=-2t
    P=e^-2t

  • Calculus -

    thank you!!

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