Finding strict Liapunov function Find a strict Liapunov function for the equilibrium point (0, 0) of
$$x'   = −2x − y^ 2$$
$$y'  = −y − x^2 .$$
Find δ > 0 as large as possible so that the open disk of radius δ and center (0, 0) is contained in the basin of (0, 0).
My attempt was $V = \frac{ax^2+by^2}{2}$, which gives
$$ \dot{V}= -2ax^2 -by^2 -axy^2 - bx^2y$$
But I couldnt find the open set. Any hint?
Thanks! 

Write $\dot{V}$ in this form:
$$\dot{V}=-(2a+by)x^2-(ax+b)y^2$$
For it to be less than $0$, we need
$$2a+by>0\implies y>-\frac{2a}{b}\\
ax+b>0\implies x>-\frac{b}{a}$$
So the disc has radius the minimum of $\frac{2a}{b}$ and $\frac{b}{a}$, or $2r$ and $\frac{1}{r}$, with $r=\frac{a}{b}$. Can you continue from here?
