- #1
MisterMan
- 47
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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.
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.