An application with partial fractions and separable equations

In summary, the model dN/dt = kN(100-N) is a good model for the spread of a highly contagious disease because it takes into account the current number of infected individuals and the potential for further spread. By using partial fractions and solving for the constant k, the function N(t) can be found to be (N + 100 - N = e^(398t + C)). However, there was an error in the last step of the calculation where the N terms should not have cancelled out.
  • #1
zeller87
1
0

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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  • #2
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.
 

FAQ: An application with partial fractions and separable equations

1. What is the purpose of using partial fractions and separable equations in an application?

Partial fractions and separable equations are used to simplify complex mathematical expressions and equations, making them easier to solve and analyze. They can also help to identify patterns and relationships between variables.

2. How do partial fractions and separable equations differ from other methods of solving equations?

Partial fractions and separable equations involve breaking down an equation into smaller, simpler parts, whereas other methods may involve direct substitution or manipulation of the entire equation. These methods are particularly useful for solving equations involving fractions or differential equations.

3. What are some common applications of partial fractions and separable equations in science and engineering?

Partial fractions and separable equations are used in a variety of fields, including physics, chemistry, and engineering. They can be applied to analyze systems such as circuits, fluids, and chemical reactions, as well as to model natural phenomena such as population growth and radioactive decay.

4. Are there any limitations or drawbacks to using partial fractions and separable equations?

While partial fractions and separable equations can be very useful, they may not always provide the most accurate or precise solutions. In some cases, other methods may be more appropriate or necessary to fully understand and solve a problem.

5. What are some tips for effectively using partial fractions and separable equations in an application?

It is important to have a clear understanding of the underlying principles and techniques involved in partial fractions and separable equations. It can also be helpful to practice with a variety of problems and to consult with others who have experience in using these methods.

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