Question #19983

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Answer 1 · 2017-06-23T21:37:55

$P = - \frac{C e^{k M t}}{1 - C e^{k M t}}$ $P=-(Ce^(kMt))/(1-Ce^(kMt))$

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

$\frac{1}{k P (M - P)} d P = d t$ $1/(kP(M-P)) dP = dt$

Integrate both sides of this equation

$\int [\frac{1}{k P (M - P)}] d P = \int d t$ $int[1/(kP(M-P))]dP =int dt$

$\frac{ln (P) - ln (P - M)}{k M} = t + C$ $(ln(P)-ln(P-M))/(kM)=t+C$

$ln (\frac{P}{P - M}) = k M t + C$ $ln(P/(P-M))=kMt + C$

Exponentiate both sides with $e$ $e$

$e^{ln (\frac{P}{P - M})} = e^{k M t + C}$ $e^(ln(P/(P-M)))=e^(kMt + C)$

$\frac{P}{P - M} = C e^{k M t}$ $P/(P-M)=Ce^(kMt)$

$P = C e^{k M t} (P - M)$ $P=Ce^(kMt)(P-M)$

$P = P C e^{k M t} - C e^{k M t}$ $P=PCe^(kMt)-Ce^(kMt)$

$P - P C e^{k M t} = - C e^{k M t}$ $P-PCe^(kMt)=-Ce^(kMt)$

$P (1 - C e^{k M t}) = - C e^{k M t}$ $P(1-Ce^(kMt))=-Ce^(kMt)$

$P = - \frac{C e^{k M t}}{1 - C e^{k M t}}$ $P=-(Ce^(kMt))/(1-Ce^(kMt))$

Answer 2 · 2017-06-23T21:38:22

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