The problem is derived from: Original gradient index optics problem
See the Figure above.
$O:(0,0)$ is the disk center of light source $\odot{O}$ with radius $3$.
Then the profile light rays of disk $O$ from the view point $B:(-14,0)$ is defined by segments $DB$ and $EB$ (also the tangent lines of $\odot{O}$ through $B$) when the refraction index is a constant value everywhere. -- By saying profile I mean, $D$ and $E$ are exactly on the edge of the visible disk.
Now if the refraction index is defined as:
$$n(x,y)=\dfrac{e^{\tfrac{(x+15)^2+y^2+12}{(x+15)^2+y^2+11}}}{e}$$
How to determine the two profile light curves of disk $O$ from the viewpoint $B$?
Per Fermat's principle, I tried to establish:
$$\delta\int{n(x,y)}\rm{d}s=0$$ and the second order nonlinear ODE via Euler-Lagrange equation: $$y''(x)=\dfrac{2\left((x+15)y'(x)-y(x)\right)\left(y'(x)^2+1\right)}{\left(y(x)^2+x(x+30)+236\right)^2}$$
In order to obtain a numerical solution, it is easy to have one boundary value condition $y(-14)=0$ of the ordinary differential equation, but I don't know how to establish another, e.g, $y'(-14)=?$.
Update
I tried some calculation, and it seems, for any numerical solution $y(x)$, it will be difficult to determine whether it is tangent to the circle $O$ or not:
