Modeling Population Growth: Solving for P(t) Using Integration

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SUMMARY

The population growth model is described by the differential equation dP/dt = 0.0004P(P-150). To derive the population function P(t), the equation is rearranged to dt = 1/(0.0004P(P-150)) dP. Integration of both sides is necessary, where the left side integrates to t + C. The right side requires the use of partial fractions to simplify the integral of 1/(P(P-150)), leading to the expression 250∫(1/P + 1/(P-150)) dP for further evaluation.

PREREQUISITES
  • Understanding of differential equations
  • Knowledge of integration techniques, specifically partial fractions
  • Familiarity with population dynamics modeling
  • Basic calculus concepts
NEXT STEPS
  • Study the method of partial fractions in integration
  • Learn about solving first-order differential equations
  • Explore applications of population models in real-world scenarios
  • Review integration techniques for rational functions
USEFUL FOR

Students in calculus or differential equations courses, educators teaching mathematical modeling, and researchers interested in population dynamics.

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Homework Statement


a population P(t) is modeled by the equation dP/dt = 0.0004P(P-150), Find a formula which gives the population, P(t), at a general time t.

Homework Equations





The Attempt at a Solution



swapping over

dt=1/0.004P(P-150) dP
then i integrate both sides

dt becomes t+c but I'm i'm not sure how to integrate the the dP side, any pointers please?
 
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Are you able to integrate \int\frac{dx}{x}?
 
First, 1/0.004= 250. You have
250\int\frac{dP}{P(P-250)}dP

Use "partial fractions". Write
\frac{1}{P(P-250)}= \frac{A}{P}+ \frac{B}{P- 250}
Find A and B and do it as two separate integrals.
 

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