Consider the autonomous DE xʹ = x(x − 1)(x+ 2). Determine the
critical points of the equation. Discuss a way of obtaining a phase portrait
of the equation.
x′=x(x−1)(x+2)Critical points:x′=0⇒x(x−1)(x+2)=0⇒⇒x∈{−2,0,1}Linearization around the critical points:x=−2:x′=6(x+2)⇒x+2=Ce6t−the trajectory runs away from −2x=0:x′=−2x⇒x=Ce−2t−the trajectory runs into 0x=1:x′=2(x−1)⇒x−1=Ce2t−the trajectory runs away from 1x'=x(x-1)(x+2)\\Critical\,\,points:\\x'=0\Rightarrow x\left( x-1 \right) \left( x+2 \right) =0\Rightarrow \\\Rightarrow x\in \left\{ -2,0,1 \right\} \\Linearization\,\,around\,\,the\,\,critical\,\,points:\\x=-2:\\x'=6\left( x+2 \right) \Rightarrow x+2=Ce^{6t}-the\,\,trajectory\,\,runs\,\,away\,\,from\,\,-2\\x=0:\\x'=-2x\Rightarrow x=Ce^{-2t}-the\,\,trajectory\,\,runs\,\,into\,\,0\\x=1:\\x'=2\left( x-1 \right) \Rightarrow x-1=Ce^{2t}-the\,\,trajectory\,\,runs\,\,away\,\,from\,\,1x′=x(x−1)(x+2)Criticalpoints:x′=0⇒x(x−1)(x+2)=0⇒⇒x∈{−2,0,1}Linearizationaroundthecriticalpoints:x=−2:x′=6(x+2)⇒x+2=Ce6t−thetrajectoryrunsawayfrom−2x=0:x′=−2x⇒x=Ce−2t−thetrajectoryrunsinto0x=1:x′=2(x−1)⇒x−1=Ce2t−thetrajectoryrunsawayfrom1
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