Modeling Population Growth: Solving a Nonlinear Differential Equation

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Discussion Overview

The discussion revolves around solving a nonlinear differential equation related to population growth, specifically the equation (5000)dN/dt = N(500-N). Participants explore integration techniques and methods for solving the equation, including partial fraction decomposition.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in solving the differential equation and questions the integration process, noting that the differential is not on top of the fraction.
  • Another participant suggests that the equation is separable and proposes separating variables to facilitate integration.
  • There is a discussion about using partial fractions to simplify the integration process, with one participant confirming that this method works.
  • Participants reflect on the tendency to overlook simpler solutions while searching for complex ones.

Areas of Agreement / Disagreement

Participants generally agree on the use of partial fraction decomposition as a solution method, but there is no consensus on the initial approach to the integration process.

Contextual Notes

Some participants express uncertainty about the integration steps and the application of the chain rule, indicating potential gaps in understanding the mathematical techniques involved.

Gilgalad
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This is out of a P3 (OCR) textbook so it should be dead easy but I just can't see the answer. I know it is not a simple ln(N(500-N)) because the differential is not on top. Anyway here it is:

(5000)dN/dt=N(500-N)

they also say when N = 100, dN/dt = 8 (which is obvious)

the answer in the back of the book is:

t=(10)ln(N/(500-N))+k

Cheers
 
Last edited:
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Hmm, most people here won't know about the exam boards we have :smile:. Have you done anything on the question and if so, could you post it?

Edit: what did you mean by the differential's not on top?

I'm stuck trying to integrate [tex]\int \frac{dN}{N(500 - N)}[/tex]. Argh :/.
 
Last edited:
The numerator

This is a variable seperable right?

so you take the N(500-N) to the left side, split the dN and the dt to get:

int[ 1/{N(500-N)} ] dN = int[ 1/5000 ] dt

Now by chain rule (I think), if the differential is on the top of the fraction you can just say it is the log of the denominator. eg int [1/x] dx = ln|x|+c

However, the differential is not on the top in this case so I am screwed.
 
Partial fractions will work.
 
oh yeah sometimes you can be so obsessed looking for something really complicated and miss something simple. It does work. Thanx
 
break the numerator into 2 pieces. I don't remember how you say that in ze english, partial fraction decomposition or something.
 
Don't worry sorted. You just needed to say partial fractions. Thanx anyway.
 

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