ModeloDeCompetencia::usage="Consider the competition model defined by: = x(1 - 0.1x - 0.05y) = y(1.7 - 0.1y - 0.15x) where the popoulations x(t) and y(t) they are measured in thousands and t in years. Analyze populations over a long period of time for each of the following cases: a x(0) = 1, y(0) = 1 b x(0) = 4, y(0) = 10 c x(0) = 9, y(0) = 4 d x(0) = 5.5, y(0) = 3.5 References: Zill, D.G.and Cullen, M.R.Differential Equations with Boundary-Value Problems. Ed. Brooks/Cole: p.89, 1997; Personal";
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s=NDSolve[{x'[t]==x[t](1-0.1x[t]-0.05y[t]),y'[t]==y[t](1.7-0.1y[t]-0.15x[t]),x[0]==1,y[0]==1},{x,y},{t,0,1000}]
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xInterpolatingFunction,yInterpolatingFunction
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Plot[Evaluate[x[t]]/.s,{t,0,1000}]
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Plot[Evaluate[y[t]]/.s,{t,0,1000}]
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s1=NDSolve[{x'[t]==x[t](1-0.1x[t]-0.05y[t]),y'[t]==y[t](1.7-0.1y[t]-0.15x[t]),x[0]==4,y[0]==10},{x,y},{t,0,1000}]
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xInterpolatingFunction,yInterpolatingFunction
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Plot[Evaluate[x[t]]/.s1,{t,0,1000}]
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Plot[Evaluate[y[t]]/.s1,{t,0,1000}]
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s2=NDSolve[{x'[t]==x[t](1-0.1x[t]-0.05y[t]),y'[t]==y[t](1.7-0.1y[t]-0.15x[t]),x[0]==9,y[0]==4},{x,y},{t,0,1000}]
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xInterpolatingFunction,yInterpolatingFunction
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Plot[Evaluate[x[t]]/.s2,{t,0,1000}]
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Plot[Evaluate[y[t]]/.s2,{t,0,1000}]
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s3=NDSolve[{x'[t]==x[t](1-0.1x[t]-0.05y[t]),y'[t]==y[t](1.7-0.1y[t]-0.15x[t]),x[0]==5.5,y[0]==3.5},{x,y},{t,0,1000}]
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xInterpolatingFunction,yInterpolatingFunction
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Plot[Evaluate[x[t]]/.s3,{t,0,1000}]
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Plot[Evaluate[y[t]]/.s3,{t,0,1000}]
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In all cases x(t) 6 and y(t) 8 when t ∞.