# Homework Help: Differential Equation Separable Variables

1. May 17, 2010

### MisterMan

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

$$\frac{dN}{dt} = k (N-n)n$$

where $$n(t)$$ 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 $$\frac{1}{[(N-n)n]}$$ as the partial fractions $$\frac{1}{N}(\frac{1}{N-n} + \frac{1}{n})$$. I manage to work this out, and I obtain this :

$$\int \frac{1}{N}(\frac{1}{N-n} + \frac{1}{n})dN = \int kdt$$

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 :

$$n = \frac{Ne^{Nkt}}{N - 1 + e^{Nkt}}$$

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

$$\int (\frac{1}{N-n} + \frac{1}{n})dN = \int Nkdt$$

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.

2. May 17, 2010

### Cyosis

First I assume $$\frac{dN}{dt} = k (N-n)n$$ should be $$\frac{dn}{dt} = k (N-n)n$$

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

$$\int (\frac{1}{N-n} + \frac{1}{n})dN = \int Nkdt$$

You are making a mess when it comes to N and n. One is a constant the other is a function.

Last edited: May 17, 2010
3. May 17, 2010

### MisterMan

N is the number of individuals in the population.

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 $$\frac{dn}{dt}$$?

4. May 17, 2010

### Staff: Mentor

I believe that the original differential equation should be
$$\frac{dn}{dt} = k (N-n)n$$

In this equation N and k are constants, and n = n(t) is the number of infected individuals.

5. May 17, 2010

### MisterMan

Thanks Mark and Cyosis, you two have been a terrific help! I did what you both suggested and used $$\frac{dn}{dt}$$ 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 , Thanks again.