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I need to prove that the system of differential equations $$ \dot x = y \\ \dot y = 1+x^2-(1-x)y $$ doesn't contain periodic solutions. I know the Bendixon criteria (that is to have div no sign changing), but, the divergence of this system is $x-1$, and I can´t sure that this will never change the sign (at $x=1$ there will be a change). How can I do that?

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  • $\begingroup$ I find periodic solutions for $x=1$ and $y<-2$. $\endgroup$ – user137035 Oct 31 '14 at 12:19
  • $\begingroup$ No, false alarm it doesn't have periodic solutions for $x=1$ and $y<-2$. $\endgroup$ – user137035 Oct 31 '14 at 12:43
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Fr any solution, $x$ is unbounded. I will use $x'$ for $\dot x$. $$ x''=y'=1+x^2-(1-x)y=1+x^2-(1-x)x'\implies x''+(1-x)x'=1+x^2\ge1. $$ Integrating $$ x'+x-\frac{x^2}{2}\ge t+c_1 \text{ for some constant }c_1. $$ Then $$ x'+x\ge t+c_1+\frac{x^2}{2}\ge t+c_1. $$ Multiplying by $e^t$: $$ (e^t\,x)'\ge(t+c_1)\,e^t. $$ Integrate and obtain that $x(t)$ grows at least as $t^2/2$ as $t\to\infty$.

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