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(1/2)y'' + y^2 = 1

I'm trying to find an approximate solution for the trajectory of a ball acted on by constant gravity acceleration and quadratic drag (proportional to the speed squared), and it turns out the speed squared satisfies an equation that is equivalent to this one after non-dimensionalizing. Any kind of solution would be nice, whether it is good for large x, small x, or whatever.

I was thinking that for x << 1, the equation would be approximately (1/2)y'' + y^2 = 0, in which case y = -3/x^2 is a solution. But then I don't know how to proceed from there to a better approximation. For x >> 1, I was thinking y^2 = 1 would be a first approximation, but again I don't know how to get any farther than that.

thanks