Differential Equation Separable Variables

In summary, the question was asking for an explicit solution for the number of infected individuals after a time t days, where n(t) is the number of infected individuals after a time t days. After some work, the person was able to find the answer using the equation n = \frac{Ne^{Nkt}}{N - 1 + e^{Nkt}}.
  • #1
MisterMan
47
0
Hi all, I have a first order separable differential equation that I find a little difficult to solve. The question is from an old maths exam paper from my country. The book I obtained it from only has the answer, but I'd really like to know how to obtain the correct general solution. Okay, here is the question :

1. A large population of N individuals contains, at time t = 0, just one individual with a contagious disease. Assume that the spread of the disease is governed by the equation

[tex]\frac{dN}{dt} = k (N-n)n[/tex]



where [tex] n(t) [/tex] is the the number of infected individuals after a time t days and k is a constant.


a) Find n explicitly as a function of t.

Part a) gives me a hint telling me to rewrite [tex]\frac{1}{[(N-n)n]}[/tex] as the partial fractions [tex]\frac{1}{N}(\frac{1}{N-n} + \frac{1}{n})[/tex]. I manage to work this out, and I obtain this :

[tex]\int \frac{1}{N}(\frac{1}{N-n} + \frac{1}{n})dN = \int kdt[/tex]

But I got stuck here as I'm not sure how to go about this as all attempts give me more than one form of n, one of n and one of e to the power of n. I had a look at the back of the book and the answer gives :

[tex]n = \frac{Ne^{Nkt}}{N - 1 + e^{Nkt}}[/tex]

So this gave me some help, so in the end I had this :

[tex]\int (\frac{1}{N-n} + \frac{1}{n})dN = \int Nkdt[/tex]

But I am not sure how to get just one n term because I don't get anywhere near the result that I am looking for when I integrate both sides.

I appreciate any helpful advice on this. Thanks in advance for any correspondence.
 
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  • #2
First I assume [tex]\frac{dN}{dt} = k (N-n)n[/tex] should be [tex]\frac{dn}{dt} = k (N-n)n[/tex]

What is N, and why are you differentiating and integrating over N? From your problem statement N seems to be a constant.

[tex]
\int (\frac{1}{N-n} + \frac{1}{n})dN = \int Nkdt
[/tex]



But I am not sure how to get just one n term because I don't get anywhere near the result that I am looking for when I integrate both sides.

You are making a mess when it comes to N and n. One is a constant the other is a function.
 
Last edited:
  • #3
Cyosis said:
What is N, and why are you differentiating and integrating over N?

N is the number of individuals in the population.

Cyosis said:
First I assume [tex]\frac{dN}{dt} = k (N-n)n[/tex] should be [tex]\frac{dn}{dt} = k (N-n)n[/tex]

No, I took that right from the book, this is what is confusing me, should I assume the book is wrong and it should be [tex]\frac{dn}{dt}[/tex]?
 
  • #4
I believe that the original differential equation should be
[tex]\frac{dn}{dt} = k (N-n)n[/tex]

In this equation N and k are constants, and n = n(t) is the number of infected individuals.
 
  • #5
Thanks Mark and Cyosis, you two have been a terrific help! I did what you both suggested and used [tex]\frac{dn}{dt}[/tex] this time, and it gave me the correct answer. I really appreciate the help, I was so frustrated that I couldn't solve this question but now that it has been explained to me that N is a constant it makes so much sense. I'm away to change the case of that n now :smile:, Thanks again.
 

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model various physical phenomena in fields such as physics, engineering, and economics.

What are separable variables in a differential equation?

In a differential equation, separable variables refer to a specific type of equation where the dependent and independent variables can be separated on opposite sides of the equation. This allows for the use of a technique called separation of variables to solve the equation.

How do you solve a differential equation with separable variables?

To solve a differential equation with separable variables, you first separate the variables on opposite sides of the equation. Then, you integrate both sides with respect to each variable and add an arbitrary constant to the solution. Finally, you can solve for the arbitrary constant using initial conditions or boundary conditions.

What are some real-life applications of differential equation separable variables?

Differential equation separable variables have various real-life applications, including in physics, chemistry, biology, economics, and engineering. For example, they can be used to model population growth, chemical reactions, and electrical circuits.

What is the benefit of using separable variables in a differential equation?

The benefit of using separable variables in a differential equation is that it simplifies the problem and allows for an easier and more straightforward solution. It also allows for the use of analytical techniques, rather than numerical methods, to solve the equation.

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