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Differential Equations

dP/dt=kP((10^6)-P). In the last 10 years the pop has grown from 100,000 to 400,000.
Determine k? k>0
(A step by step method will be really helpful, I have past the process of intergrating both sides and have solved the partial fractions).

Reply 1

BabyMaths

Original post by Engineeringbeast

People will be more likely to help if you post what you've done.

Reply 2

Solivagant

Original post by BabyMaths

People will be more likely to help if you post what you've done.

And post in LaTeX.

Reply 3

CTArsenal

Engineeringbeast you really have to get used to Latex; this question doesn't need it much but some old questions you've had aren't understandable unless you use it haha. Here's a link for reference: http://www.thestudentroom.co.uk/wiki/latex

This is the question in Latex (well at least I hope it is):

\frac{dP}{dt} = {kP(10^6-P)}

You should just get:

\int k\ dt = \int \frac{1}{P(10^6-P)}\ dP

Once you've solved it, you just sub in your P value of 100,000 at t = 0 (as that's the initial population) to get the constant integer, then sub in your P value of 400,000 at t = 10. This is assuming t is measured in years and P is measured per person.