Integrating for logistic growth model

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SUMMARY

The discussion focuses on solving the differential equation for the logistic growth model represented by M'(t) = M(S-M) + I, where M(t) indicates biomass growth and I signifies immigration. The user, Darkmisc, attempts to solve this equation using separable differential equations and partial fractions but encounters difficulties, particularly with the term involving immigration. The solution using Wolfram Mathematica does not isolate M, prompting a request for clarification on the application of partial fractions in deriving the general logistic equation.

PREREQUISITES
  • Understanding of differential equations, specifically separable equations.
  • Familiarity with the logistic growth model and its components.
  • Knowledge of partial fraction decomposition techniques.
  • Experience with mathematical software, such as Wolfram Mathematica.
NEXT STEPS
  • Study the application of partial fractions in solving differential equations.
  • Learn how to isolate variables in differential equations using Wolfram Mathematica.
  • Research the derivation of the general logistic equation from first principles.
  • Explore advanced techniques in differential equations, such as integrating factors.
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Students studying differential equations, mathematicians interested in logistic growth models, and anyone seeking to understand the application of partial fractions in mathematical modeling.

Darkmisc
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Homework Statement



Hi, can anyone help me solve a differential equation for the logistic growth model?

Homework Equations





It reads:

M'(t) = M(S-M) + I, where M(t) represents the growth of a biomass. "I" represents immigration (in a coral reef) and there is no breeding.




The Attempt at a Solution



I've treated it as a separable differential equation, but get the term

(integral of) 1/(M(S-M)+I) dM

A solution is possible using wolfram mathematica, but it doesn't isolate M.

I'd solve the equation using partial fractions, but the "I" term seems to make it impossible. Am I right in saying that?

Also, going back a step, can anyone explain how to use partial fractions to derive the general logistic equation. Wikipedia has a section on it: http://en.wikipedia.org/wiki/Partial_fraction

but there's a step I don't understand, namely: A=B, A=1/M, B=1/M. On what basis was this assumed?

In trying to solve it myself, I let A=0 and got B= 1/P, then let B=0 and got A=1/(M-P).

When these figures for A and B are substituted back into the original equation, I get 2/(P(M-P)) .

Can anyone explain what I'm doing wrong?


Thanks,

Darkmisc


 
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Darkmisc said:
Also, going back a step, can anyone explain how to use partial fractions to derive the general logistic equation. Wikipedia has a section on it: http://en.wikipedia.org/wiki/Partial_fraction

but there's a step I don't understand, namely: A=B, A=1/M, B=1/M. On what basis was this assumed?

In trying to solve it myself, I let A=0 and got B= 1/P, then let B=0 and got A=1/(M-P).

When these figures for A and B are substituted back into the original equation, I get 2/(P(M-P)) .

Can anyone explain what I'm doing wrong?

Say you are trying to set up this partial fraction expansion and figure out A and B:

\frac 1 {m(s-m)} = \frac A m+ \frac B {s-m}

Add the two fractions on the right to get:

\frac 1 {m(s-m)} = \frac {A(s-m)+Bm}{(m)(s-m)}

For these to be equal, the numerators must be equal:

1 = A(s-m)+Bm

If you put m = 0 you get A = 1/s and if you put m =s you get B = 1/s. So your partial fraction expansion becomes:

\frac 1 {m(s-m)} = \frac {\frac 1 s} m+ \frac {\frac 1 s} {s-m}
 

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