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Check equilibrium point for stability

Writer Sophia Terry
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$\dot x = -y^2$
$\dot y = x^3y$

The task is to draw global phase portrait (I guess I understand how to manage this using isoclines) and to check equilibrium point for stability. I found out that equilibrium points are (x; 0) and using Jacobian matrix found out, that if x > 0 one of eigenvalues is always positive so points (x, 0) where x > 0 are not stable at all. (Correct me, if I am wrong, please). And I don't know how to check the solutions where x <= 0.

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1 Answer

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Note that $x^4 + 2 y^2$ is constant on trajectories of this system. The curves $x^4 + 2 y^2 = c$ are shaped like distorted ellipses. Any solution in the left half plane will approach the negative $x$ axis along one of these curves. In particular, the equilibrium points $(x,0)$ for $x \le 0$ are stable (but not asymptotically stable).

EDIT:

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