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celia
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Just in case you have the paper and you could avoid my poor attempts at computer notation, it's June 99, OCR MEI. q3

50 (df/dt) = r

(dr/dt) = r - 8f

where f is the fow population and r is the rabbit population at time t years after the foxes are introduced.

i) (I can do this) show that 25 (d2r/dt2) - 25 (dr/dt) + 4 r = 0

ii) solve this to find the general solution for r (can do this I think, it's r = Ae^20t + Be^5t )

Initially, r = 12000 and f = F

find the solutions for f and r subject to these conditions.

This is where it may all be going wrong, although I think

r = (1500 + 8/15 F) e^5t - (3200 + 8/15 F) e^20t

f = (375/2 + 1/15 F) e^5t - (400 + 1/15 F) e^20t

iv) Show that for F> 1200 the population of rabbits will die out.

v) Is it possible for the fox population to die out before the rabbit population? Explain your answer.


Thank you very much to anyone who has had the patience to read through all this, and especially if you've been able to help!
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Jonny W
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(Original post by celia)
Just in case you have the paper and you could avoid my poor attempts at computer notation, it's June 99, OCR MEI. q3

50 (dr/dt) = r

(dr/dt) = r - 8f
I think there is a mistake here. Should the first equation involve df/dt?
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minkailin
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this question is in the MEI M4 textbook, answers in the back.
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Fermat
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(Original post by Jonny W)
I think there is a mistake here. Should the first equation involve df/dt?
Yeah, it should be df/dt -just seen the answer scheme!
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Fermat
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Check your solution to the DE

The auxiliary eqn is,

25m² - 25m + 4 = 0
m = [25 ±√(625 - 400)]/50
m = [25 ±√(225)]/50
m = ½ ±√(9)/10
m = ½ ±3/10
m = 4/5 , m= 1/5

r = Ae^(4t/5) + Be^(t/5)
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