Modeling Population Growth: Solving a Nonlinear Differential Equation

[Gilgalad]

This is out of a P3 (OCR) textbook so it should be dead easy but I just can't see the answer. I know it is not a simple ln(N(500-N)) because the differential is not on top. Anyway here it is:

(5000)dN/dt=N(500-N)

they also say when N = 100, dN/dt = 8 (which is obvious)

the answer in the back of the book is:

t=(10)ln(N/(500-N))+k

Cheers

 

[Nylex]

Hmm, most people here won't know about the exam boards we have . Have you done anything on the question and if so, could you post it?

Edit: what did you mean by the differential's not on top?

I'm stuck trying to integrate \int \frac{dN}{N(500 - N)}. Argh :/.

 

[Gilgalad]

The numerator

This is a variable seperable right?

so you take the N(500-N) to the left side, split the dN and the dt to get:

int[ 1/{N(500-N)} ] dN = int[ 1/5000 ] dt

Now by chain rule (I think), if the differential is on the top of the fraction you can just say it is the log of the denominator. eg int [1/x] dx = ln|x|+c

However, the differential is not on the top in this case so I am screwed.

 

[shmoe]

Partial fractions will work.

 

[Gilgalad]

oh yeah sometimes you can be so obsessed looking for something really complicated and miss something simple. It does work. Thanx

 

[inha]

break the numerator into 2 pieces. I don't remember how you say that in ze english, partial fraction decomposition or something.

 

[Gilgalad]

Don't worry sorted. You just needed to say partial fractions. Thanx anyway.