Solving the Gompertz Differential Equation

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SUMMARY

The discussion focuses on solving the Gompertz differential equation (DE) with specific parameters. The participants analyze cases where a = b = 1 and a = 1, b = -1, and utilize phase portraits to sketch solution curves for initial conditions P0 > e and 0 < P0 < e. The method of separation of variables is employed to derive an explicit solution, leading to the expression P = e^(Ae^(-bt) - a)/-b. The conversation emphasizes the importance of understanding the phase portrait concept and integration techniques in solving differential equations.

PREREQUISITES
  • Understanding of differential equations, specifically the Gompertz model.
  • Familiarity with phase portrait analysis in dynamical systems.
  • Proficiency in integration techniques, including separation of variables.
  • Knowledge of logarithmic functions and their properties.
NEXT STEPS
  • Study the phase portrait method for analyzing autonomous differential equations.
  • Explore advanced integration techniques relevant to differential equations.
  • Learn about the applications of the Gompertz model in population dynamics.
  • Investigate numerical methods for solving differential equations when analytical solutions are complex.
USEFUL FOR

Students and researchers in mathematics, particularly those studying differential equations and their applications in biological modeling, as well as educators teaching these concepts.

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Homework Statement


The problem in the book:

a) Suppose a = b = 1 in the Gompertz differential equation. Since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch representative solution curves corresponding to the cases P0 > e and 0 < P0 < e.

b) Suppose a = 1, b = -1 in the Gompertz DE. Use a new phase portrait to sketch representative solution curves corresponding to the cases P0 > e-1 and 0 < P0 < e-1

c) Find an explicit solution of the Gompertz DE subject to P(0) = P0

Homework Equations


dP/dt = P(a-blnP)

The Attempt at a Solution


I used separation of variables to get:
dP/(P(a-blnP)) = dt

I let u = a - blnP and du = -bdP/P which leaves me with:
-b\intdu/u = \intdt

I integrate to get:
-b ln (u) = t + C

ln (u-b) = t + C

eln (u-b) = et + C

u-b = Aet

(a - b ln P)-b = Aet

So how do I solve for P? :P Or, am I even close to having the right answer? LOL
 
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P = eAe-t/b?
 
Last edited:
Or how about:

P = e(Ae-bt - a)/-b
 

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