buckmo
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Problem:
x = number of rabbits
y = number of foxes
Accept the usual assumptions involved in predator-prey (Lotka-Volterra) problems.
dx/dt = 0.04x - 0.002xy
dy/dt = -0.08y + 0.0004xy
Initial conditions: 200 rabbits and 10 foxes
This problem is in a textbook. A graph of the functions x and y when 0 </= t </= 240 is illustrated without showing the functions that are graphed. I am trying to generate those functions.
I am having trouble solving the differential equations. I began by separating variables...
dx/dt = 0.04x - 0.002xy
500(1/x)dx = (20 - y)dt
500 ln(x) = t(20 - y) + C1
Using initial conditions, C1 = 500 ln(200), so
500 ln(x) = t(20 - y) + 500 ln(200)
Solving the other equation in a similar way gives
2500 ln(y) = t(x - 200) + 2500 ln(10)
From here I tried several techniques put found myself painted into a corner. I am thinking I need to rethink the method of solving the equations above. Euler's Method may work to approximate values, but I don't know how to set it up. Suggestions?
Thanks for taking a look.
x = number of rabbits
y = number of foxes
Accept the usual assumptions involved in predator-prey (Lotka-Volterra) problems.
dx/dt = 0.04x - 0.002xy
dy/dt = -0.08y + 0.0004xy
Initial conditions: 200 rabbits and 10 foxes
This problem is in a textbook. A graph of the functions x and y when 0 </= t </= 240 is illustrated without showing the functions that are graphed. I am trying to generate those functions.
I am having trouble solving the differential equations. I began by separating variables...
dx/dt = 0.04x - 0.002xy
500(1/x)dx = (20 - y)dt
500 ln(x) = t(20 - y) + C1
Using initial conditions, C1 = 500 ln(200), so
500 ln(x) = t(20 - y) + 500 ln(200)
Solving the other equation in a similar way gives
2500 ln(y) = t(x - 200) + 2500 ln(10)
From here I tried several techniques put found myself painted into a corner. I am thinking I need to rethink the method of solving the equations above. Euler's Method may work to approximate values, but I don't know how to set it up. Suggestions?
Thanks for taking a look.