Question #19983

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Answer 1 · 2017-06-23T21:37:55

$P=-(Ce^(kMt))/(1-Ce^(kMt))$

Assuming the $M$ is a constant, you would need to first separate the variables so that

$1/(kP(M-P)) dP = dt$

Integrate both sides of this equation

$int[1/(kP(M-P))]dP =int dt$

$(ln(P)-ln(P-M))/(kM)=t+C$

$ln(P/(P-M))=kMt + C$

Exponentiate both sides with $e$

$e^(ln(P/(P-M)))=e^(kMt + C)$

$P/(P-M)=Ce^(kMt)$

$P=Ce^(kMt)(P-M)$

$P=PCe^(kMt)-Ce^(kMt)$

$P-PCe^(kMt)=-Ce^(kMt)$

$P(1-Ce^(kMt))=-Ce^(kMt)$

$P=-(Ce^(kMt))/(1-Ce^(kMt))$

Answer 2 · 2017-06-23T21:38:22

Please see