Finding periodic and heteroclinic orbits?

Question

Does the following system have any periodic or heteroclinic orbits?

$ \frac{dx}{dt} = -x $

$ \frac{dy}{dt} = y - y^4 $

So if a periodic orbit exists, the solution x(t) = x(t + a) for any a.

For the x part i get x(t) = $ Ae^{-t} $ which doesn't repeat itself in time.

Is this the right technique and if so how do i do the y part? (i don't think im meant to solve it as it's quite a complicated integral and it's only a 3 mark question)

For the heteroclinic orbit i found the equilibrium points (0,0) and (0,1).

and using the above solution for the x part, the solution doesn't tend to $ e^{-1} $ as t goes to infinity or negative infinity and so i concluded that the solution has no heteroclininc orbit. Is this correct?

Thanks!

Answer 1

This is an old question, but anyways:

$1)$ There are no periodic orbits since at least $x'=-x$ has no periodic solutions. Nothing else is needed here.

$2)$ The equilibria are indeed $(0,0)$ and $(0,1)$. There is a heteroclinic orbit between the two (the vertical segment between the two) as a consequence of the phase portrait of $y'=y(1-y^3)$.