diff. equations: population growing logistically (flies)

[Hansel13]

A population of flies choose to reproduce logistically. The initial population is 1000, and growing at the (net) instantaneous rate of 50 flies a day.

What is the explicit formula which represents the fly population as a function of time?

So I got this...
dP/dt = 50P*(1-(P/10,000))
And I used the separable equation method to find the wanted function. But I got the problem wrong, and my teacher said that K=1/18, not 50.

Can someone tell me how he got K to equal 1/18? I'm lost.

 

[skeeter]

Re: diff. equation problem

assuming that the max is 10000 ...

dP/dt = kP(1 - P/10000)

at t = 0, P = 1000 and dP/dt = 50

50 = k*1000(1 - 1000/10000)

50 = k*1000(1 - .1)

50 = k*1000(.9)

900k = 50

k = 1/18

 

[Hansel13]

Re: diff. equation problem

assuming that the max is 10000 ...

dP/dt = kP(1 - P/10000)

at t = 0, P = 1000 and dP/dt = 50

50 = k*1000(1 - 1000/10000)

50 = k*1000(1 - .1)

50 = k*1000(.9)

900k = 50

k = 1/18

Yes, sorry, the Max is 10,000.

Thanks a lot! Now I should be able to find the solution.