- **Calculus & Beyond Homework**
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- - **Differential Equations - Logistic Model**
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Differential Equations - Logistic ModelI have the equation dP/dt = kP(1 - P/A). It is supposed to describe a logistical situatuon involving the carrying capacity of the system.
k is a constant, and A is the carrying capacity of the system. t is time and P is population as a function of time. P(0) = P _{0}. I solved c (the integration constant) to be:c = -ln|(P _{0})/(A - P_{0})|I'm trying to solve the equation in terms of t. In my calculus book, a similar equation is given with an explanation. dP/dt = 0.1P(1 - P/300) With an initial condition of P(0) = 50, c is found to be ln(1/5). A = 300 and k = 0.1. I follow along well up to this point.After solving for c, the book lists the rearranged equation as: P(t) = 300/(1 + 5e ^{-0.1t})I don't understand how they went from one equation to the other, the closest I could come with the general equation was: P(t) = (Ae ^{kt} + Q)/(1 + e^{kt}) where Q = (P_{0})/(A - P_{0})Which would coincide with an equation of: (300e ^{0.1t} + .2)/(1 + e^{0.1t})Which, when graphed, is not equivalent to the equation given by the book. Can anyone go over how to solve the general equation? I think I'm missing some crucial point.Thank you! |

Re: Differential Equations - Logistic ModelThis equation is separable - meaning that you can get all of the terms involving P on one side (including dP), all of the terms involving t on the other (including dt), and integrate both sides.
So we write the equation as [tex]\left(\frac{1}{P} + \frac{1/A}{1-P/A}\right)\ dP = k\ dt[/tex] Integrate both sides to get ln|P| - ln|1-P/A| = kt + c, take the exponential of both sides, and you'll get your answer |

Re: Differential Equations - Logistic ModelThat's what I've done. I'm trying to solve for P, and I can't figure out where my method diverges from the book's explanation.
I was hoping someone could show me the steps to solving for P, so I can figure out where my error lies. |

Re: Differential Equations - Logistic ModelWell, exp(ln|P|-ln|1-P/A|) = exp(kt + c)
The right hand side simplifies to exp(kt+c) = exp(kt)*exp(c) = C exp(kt), where C is a constant that we'll determine later. The left hand side simplifies to exp(ln|P|-ln|1-P/A|) = exp(ln|P|) exp(-ln|1-P/A|) = P/(1-P/A) So P/(1-P/A) =C exp(kt) P = (A-P) C exp(kt)/A P = AC exp(kt) - (PC/A)exp(kt) P + (PC/A) exp(kt) = AC exp(kt) P(1 + C/A exp(kt)) = AC exp(kt) P = AC exp(kt)/(1+C/A exp(kt)) Multiplying the numerator and denominator of the RHS by exp(-kt) P = AC/(exp(-kt) + C/A) |

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