# The zero point $(0,0)$ of $x'=x+y,y'=xy-bx-y$ is stable when $b>1$

Proof when $b>1$, the zero point $(0,0)$ of ODE \left\{\begin{align}&x'=x+y\\&y'=xy-bx-y\end{align}\right. is stable.

I couldn't find a proper Lyapunov V function.

Let $v=x'$ we has 1st order ODE $v\frac{dv}{dx}=xv-x^2-bx$, but it has no suitable expressions

• Did you find the critical points, the Jacobian, and then analyze the Jacobian at each critical point for the parameter b? – Moo Jul 2 '16 at 5:58
• @Moo It has zero real part and non-zero imag part. – yaoliding Jul 2 '16 at 6:32
• Related (see @Artem's detailed solution), here the perturbation term is $xy$. – Did Jul 2 '16 at 8:55