Determine the stability of $(x,y)=(0,0)$ Determine the stability of $(x,y)=(0,0)$:
1/$$\bf{\begin{cases}
 & \mathrm{  } \dot{x}= -2x-y+2xy^2-3x^3\\ 
 & \mathrm{  } \dot{y}= \dfrac{x}{3}-y-x^2y-7y^3
\end{cases} \tag {1}}$$
2/ 
$$\bf{\begin{cases}
 & \mathrm{  } \dot{x}= x-xy^4\\ 
 & \mathrm{  } \dot{y}= y-x^2y^3
\end{cases} \tag {2}}$$
I'm trying to find a Liapunov function $V(x,y)=???$. But I have no solution! :( Anyone can post a few hints (like $V(x,y)=?$) 
Any help will be appreciated.Thanks!
 A: Hints:
If we look at phase portraits for the two systems, we have:


From these, we can see there is a single critical point for the first and it looks well behaved, so there is a chance we can find one.
Hint: Try $V(x, y) = ax^2 + by^2$ and see if you can resolve $a$ and $b$.
For the second one, there are a lot of things going on in the phase portrait due to five critical points. I doubt you will be able to find one for this as it is all over the map. Also, it is clear the CP is unstable by just looking at the direction fields, but you can do an eigenvalue analysis of the Jacobian at $(0,0)$ to also see that.
A: Solution: 
For system of equations: $$\bf{\begin{cases}
 & \mathrm{  } \dot{x}= x-xy^4\\ 
 & \mathrm{  } \dot{y}= y-x^2y^3
\end{cases} \tag {2}}$$
We use the Chetaev theorem to prove this.
Let the function $V(x,y)$ have the form $$V(x,y)=x^2-y^2$$
This function is positive definite in the subdomain $\mathcal{V}$ in which the inequality $|x| > |y|$ holds. 
We calculate the derivative $\dfrac{dV}{dt}$ by virtue of the system and define its sign in the subdomain $\mathcal{V}$. 
$$\dot{V}(x,y)=2(x^2-y^2)=2V(x,y)$$


*

*It can be seen that the derivative $\dfrac{dV}{dt}$ is also positive definite in the subdomain $\mathcal{V}$ defined by the relation $|x| > |y|$.

*Besides that, the function $V(x,y)$ is zero on the boundary of $\mathcal{V}$ including the point $(0,0)$. 
So, all conditions of the Chetaev theorem are fulfilled. Therefore, the zero solution of the system is unstable. 
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Is that ok? Amzoti
