An application with partial fractions and separable equations

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SUMMARY

The discussion centers on modeling the spread of a contagious disease using the differential equation dN/dt = kN(100-N), where N(t) represents the number of infected individuals in thousands. The user calculated the constant k as 3.98 by substituting known values into the equation. They attempted to solve the equation using partial fractions but encountered an error in their logarithmic manipulation, specifically in the step where they incorrectly canceled terms. The correct approach involves recognizing that ln(a) + ln(b-a) equals ln(a(b-a)), which preserves the N term.

PREREQUISITES
  • Understanding of differential equations, specifically separable equations
  • Familiarity with partial fraction decomposition
  • Knowledge of logarithmic properties and manipulations
  • Basic epidemiological modeling concepts
NEXT STEPS
  • Study the method of solving separable differential equations
  • Learn about partial fraction decomposition techniques in calculus
  • Explore the application of logarithmic identities in solving equations
  • Research epidemiological models and their parameters, such as the SIR model
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Students and professionals in mathematics, epidemiology, or public health who are interested in modeling disease spread and solving differential equations.

zeller87
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Homework Statement


Suppose that a town has a population of 100,000 people. One day it is discovered that 1200 people have a highly contagious disease. At that time the disease is spreading at a rate of 472 new infections per day. Let N(t) be the number of people (in thousands) infected on day t.

1. Explain why the model dN/dt = kN(100-N) might be a good model for the spread of the disease
2. Use partial fractions to solve the differential equation and find the function N(t).
3. According to this model if the disease continues unchecked when will the number of people infected reach 30,000.
[suggestion: solve for the constant k before you do any integration]

Homework Equations





The Attempt at a Solution


From the equation given in part 1, I plugged in the values give:

472 = k(1.2)(100-1.2)

and found k = 3.98. From there I separated the problem and integrated:

∫1/(N(100-N))dN = ∫3.98dt

using to partial fractions to solve, I found this to reduce to:

(1/100)ln|N| + (1/100)ln|100-N| = 3.98t + C
ln|N| + ln|100-N| = 398t + C
N + 100 - N = e^(398t + C)

But the N's cancel out and I can't find the mistake i have made. Any help would be great.
 
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I didn't follow all of your working step by step, I just noticed that the very last line of your calculation isn't right. ln(a) + ln(b-a) != ln(b). Rather, ln(a) + ln(b-a) = ln(a(b-a)). In other words, when you exponentiate everything on the left hand side, the N term shouldn't cancel.
 

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