So, I saw this equation as (I think) a joke somewhere online:
y'' = y'^(y'*sin(x*y)) + (e^x)*tan(1/x)
It's way too advanced for me, and I can tell the solution is probably really nasty. Can any of you solve it? Wolframalpha wasn't giving it.
One thing you can do is cast it in to an easier problem to solve numerically. Maybe something like the following
Let u = y, v=y'. Then the equation becomes
v' = v^(v*sin(x*u)) + (e^x)tan(1/x),
u' = v.
Then you can take the initial point (u_0, v_0) and expand the derivatives as finite-differences to get an Euler scheme:
v_{n+1} =v_n+h_n*(v_n^(v_n*sin(x_n*u_n))+(e^(x_n))tan(1/x_n)),
u_{n+1}= u_n + h_n*v_n,
where h_n = x_{n+1}-x_n. We almost know there is a region where we can find a solution. (ODEs are well behaved like this, but I don't have the theorems off the top of my head.) We obviously need to be away from x=0, and also there are going to be regions (1/x = (n+1/2)*Pi) where the solution blows up. So you'd probably have to stick to strips between these points. Inside of there I'd guess you could converge to a solution, though.
You don't need to reduce the order, you could also directly discretize the original problem
(y_{n+1} - 2y_n+y_{n-1})/(2h_n^2) = ((y_n-y_{n-1})/h)^((y_n-y_{n-1})/h)*sin(x_n*y_n)) + (e^x_n)*tan(1/x_n)
with an initial y_0, y_1. Basically the same as before.
Edit: I shouldn't have said initial guess; the point (u_0,v_0) = (y_0, y'_0) must be provided for the problem to have a chance of being well defined.