I am reading about Euler-Lagrange equations and this particular section is a little unclear. Consider the differential equation
$$\begin{bmatrix} \dot{x}\\ \dot{y} \end{bmatrix} = \begin{bmatrix} -7x^{\frac{6}{7}}y^{\frac{1}{7}}\\ 7x^{\frac{1}{7}}y^{\frac{6}{7}} \end{bmatrix}\quad (x(0),y(0)) = (x_{0},y_{0}), \quad x_{0},y_{0} \neq 0.$$
How do I solve this using the fact that $H(x,y) = x^{\frac{2}{7}} + y^{\frac{2}{7}}$ is a first integral of the above differential equation?
If we let $u = (x,y)$, $u(0) = (x_{0},y_{0})$ be a solution to the above differential equation, then $H(u)$ is a constant. In particular, $H(u) = H(u(0))$ for all $t$ since a first integral is constant along the solutions of the differential equation.
After this, I am a confused. I know that the level sets of $H(x,y)$ looks like an asteroid. But how does this help me? Is each level set of $H$ the orbit of a solution to the differential equation? How do I know that there aren't any extraneous points in the level set that are not part of $u$?
If the above is correct, I don't understand how to write down the general solution from this level curve.
The level sets of a first integral are not necessarily orbits, but the orbits are pieces of these level sets. If $L$ is a level set and $E$ the set of equilibrium points, the connected components of $L \backslash E$ should be orbits.
EDIT: If you can parametrize a level set (say as $x = X(s), y=Y(s)$), then substituting $x = X(s(t)), y = Y(s(t))$ in to the differential equation should give you a single first-order differential equation for $s(t)$. In your example I would try $x = c\; \cos(s)^7$, $y = c\;\sin(s)^7$.