Logistic modeling - help integrating/solving for P

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SUMMARY

The discussion focuses on solving a logistic modeling equation represented by the differential equation dP/dt = (1/900)P(9-P). The user attempts to integrate this equation using separation of variables and partial fractions, ultimately expressing the solution in terms of natural logarithms. Key steps include the integration of 1/(P(9-P)) and the use of constants to simplify the expression. The user expresses confusion regarding the growth behavior of P and the correct application of logarithmic properties.

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  • Understanding of differential equations, specifically separable equations.
  • Familiarity with logistic growth models and their mathematical representation.
  • Knowledge of integration techniques, including partial fractions and logarithmic properties.
  • Basic algebraic manipulation skills to handle constants and expressions.
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  • Study the method of separation of variables in differential equations.
  • Learn about logistic growth models and their applications in real-world scenarios.
  • Explore integration techniques, particularly focusing on partial fractions.
  • Investigate the behavior of solutions to logistic equations as time approaches infinity.
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Students studying differential equations, mathematicians interested in logistic modeling, and educators seeking to enhance their understanding of integration techniques in applied mathematics.

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Homework Statement


Sorry about the title; I accidentally hit enter instead of 'Shift'. It should read Logistic Modeling -- help integrating/solving for P

If P(0)=2, find P(90).

Homework Equations



dP/dt=1/900P(9-P)

The Attempt at a Solution



my solution looks like:

dP/(P(9-P))=1/900dt (seperable diff eq.)

1/9(ln(P)(9-P))=1/900 t+C (partial fractions and property ln M + ln N = ln MN)

lnP(9-P) = 1/100t+9C

P(9-P)= Ce^(1/100t) (exponentiated; 9C=C)

kinda stuck here (and I don't see how P ever stops growing.)
 
Last edited:
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be careful with your brackets, its pretty hard to read what is in your log, divided etc.

but following through your method
\frac{1}{P(9-P)} = \frac{A}{P}+\frac{b}{9-P} = \frac{9A + (b-A)P}{P(9-P)}
giving A = 1/9, B = 1/9

so evaulating the integral
\int \frac{dP}{P(9-P)} = \frac{1}{9}(\int\frac{dP}{P}+\int\frac{dP}{9-P}) =\frac{1}{9}(ln(P) - ln(9-P)) +C = \frac{1}{9}ln(\frac{P}{9-P})+C

maybe you missed a negative...?
 

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