Existence of periodic solution in an ODE system Let the ODE system,
\begin{cases}x'(t)=x^3+x^5 \\ y'(t)=y+y^7 \end{cases}
defined in $\mathbb{R}^2$.
I need to say if this system supports periodic solution with period $T>0$.
This is a question of Analysis in $\mathbb{R}^n$ not ODE, that is, I need to demonstrate this using concepts presented in Analysis in $\mathbb{R}^n$, my textbook is Analysis on Manifolds by Munkres. The hint is,

A periodic solution of the system above is a closed curve $\gamma(t) = (x(t), y (t))$ that satisfies the system.

I believe I have to use Stokes, but I am not able to proceed with the solution.
 A: You can consider the curve $\phi = (x, y)$ and take the squared distance to the origin $x^2 + y^2$. The derivative of this is
$$2(x\dot{x} + y\dot{y}) = 2(x^4 + x^6 + y^2 + y^8)\ge 0.$$
If this curve is periodic, then it is actually a closed curve and so a compact subset of the plane so there must be a minimum of this distance, corresponding to a point closest to the origin.
Since it is a sum of squares this means $x = y = 0$ at that point, so your curve goes through the origin, but clearly the constant curve $\phi\equiv 0$ satisfies the equation and it can be the only one doing that.
If you wanted to conclude this from Stokes Theorem you need to integrate the 1-form
$$(x^3 + x^5)dx + (y + y^7)dy$$ along the line given by a potential closed curve, i.e. around $\phi$ and if $D$ is the interior, assuming $T$ is the minimum period so that you avoid self intersections, you conclude
$$\int_C (x^3 + x^5)dx + (y + y^7)dy = \int\int_D \dfrac{d(y + y^7)}{dx} - \dfrac{d(x^3 + x^5)}{dy} dxdy = 0.$$
But, by definition of evaluating a line integral, this is the same as
$$\int_C (x^3 + x^5)dx + (y + y^7)dy = \int \phi'(t)\cdot\phi'(t) dt,$$
where we have used that $\phi$ satisfies the differential equation to get the first $\phi'$. The integrand is nonnegative and since it integrates to $0$, it is equal to $0$ everywhere in the interval it is defined. That is, $\phi \equiv 0$.
A: Hint: $x^3 + x^5 > 0$ when $x > 0$, $< 0$ when $x < 0$.  Similarly for $y + y^7$.
A: If there is a solution that is periodic in $(x(t),y(t))$, then also functions $x(t)$ and $y(t)$ have to be periodic separately, as real functions.
Further, the system is decoupled, you have two scalar, non-coupled autonomous ODE of first order. Scalar autonomous first-order ODE do not have periodic solutions, their solutions are constant or (strictly) monotonous (assuming the right side function is locally Lipschitz or differentiable).
A: The decoupled dynamic system can be integrated as follows
$$
\cases{
\frac{dx}{x^3+x^5}= dt\\
\frac{dy}{y+y^7}= dt
}\ \ \Rightarrow \cases{\ln \left(x^2+1\right)-2 \ln (x)-\frac{1}{x^2}=2t\\ \ln (y)-\frac{1}{6} \ln \left(y^6+1\right)=t}\ \ \Rightarrow \cases{\psi(x) = 2t\\ \phi(y) = t}
$$
As can be verified $\psi, \phi$ are analytic strict increasing functions in $\mathbb{R}_{+}$ so their inverses $x=\psi^{-1}(t), y=\phi^{-1}(t)$ are also strict increasing functions. The conclusion is that under such circumstances no periodic orbit is possible.
