Question

I need help with the last part of the problem please

Consider the following model for the populations of rabbits and wolves (where R is the population of rabbits and W is the population wolves). DR = 0.05R(1 -0.0005R) - 0.000703125RW dW = -0.08W +0.0004RW Find all the equilibrium solutions: In the absence of wolves, the population of rabbits approaches 2000 help (numbers) help (numbers) In the absence of rabbits, the population of wolves approaches 0 If both wolves and rabbits are present, their populations approach Rabbits = 200 . help (numbers) and Wolves - help (number)

Answer 1

Sol @ Given dt = 0.05R (1-0.0005 R) - 0.000703125 RW de er -> 0.05R (1-0.0005R) - 0.000703125 RW:0 & dW dt = -0.08W +0.0004RW --) -0.08W +0.0004 RW-O a) In the absence If W=o. of Wolves then, AR alt = 0.05R 11-0.0005R) 0.05 gives the date of initial growth, which is Exponential while the second torm (1-0.0005R) provides the limiting factos Set (1-0.0005R) = 0 130.0005R R=0.0005 R= 2000 . In the absence of sabbits, (If R=0) then. we have du = -0.08w This is exponential decay at the date 0.08 This leads to a population of 'o. (W=o]

c) Now with neither w ork cornal to o, we set the derivative coval to'o dw =-0.08W+0.0004 RW As W -0.08W+0.000 URW=0. W(-0.08 +0.0004R)=0 is not conal to o -0,08 +0.0004R=0 0.08 = 0.0004 R R=0.08 0.0004 R=200] Substitute R into the eqn @ 0.25 R (1-0.0005R)- 0.000 703125 RW=0 (0.05)(200) (1-(0.0005)(2009) - 60.000703125) (200) W=0 (0) (1-0.1) = (0.140625)W (10%0.9) = 0.140625 W 9 = 0.140625 W W= 9 0140625 W=64 (R, W) = (200,64) =o=