this is a problem from one of the former exams from ordinary differential equations: as stated in topic, find for which parameters a the solution (0,0) is asymptotically and lapunov stable.
$$ x'= (2a+1)x- \frac{y}{y+1}$$ $$ y'= sinx-(a+1)y$$ My idea is to reduce it by taylor expansion , as i have done it
$$x'= (2a+1)x-y+y^2$$ $$y'= x-x^3-(a+1)y$$ Now we get rid of the higher degree expressions to use a theorem, we get $$x'= (2a+1)x-y$$ $$y'= x-(a+1)y$$ and we calculate the soultions of its characteristic polynomial, their real part has to be below zero to make the solution stable and after calculating it we get that for $$ -3/2<a<0$$ it is asomptytically stable My question is, is it even correct?! and what can I do for others a
Here is a picture of the phase portrait for $a = -3/2$, that is, for $$ \begin{cases} x'= -2 x- \frac{y}{y+1} \\ y'= \sin{x} - \frac{y}{2}. \end{cases} $$
The origin appears to be unstable.
For the second marginal case ($a = 0$), that is, $$ \begin{cases} x'= x- \frac{y}{y+1} \\ y'= \sin{x} - y, \end{cases} $$ a picture is
Again, this looks like an unstable equilibrium.
Finally, a picture of $$ \begin{cases} x'= - x- \frac{y}{y+1} \\ y'= \sin{x} \end{cases} $$
(a stable focus),
and of
$$ \begin{cases} x'= - 1.8 \, x- \frac{y}{y+1} \\ y'= \sin{x} + 0.4 \, y \end{cases} $$
(a stable node).