1. The problem statement, all variables and given/known data Solve the reducible 2ODE. Assume x, y and/or y' positive where helpful. y^3 * y'' = 1 3. The attempt at a solution Well, I tried what I normally would do for x being missing. p = (dy/dx); y'' = p'p = (dp/dy)(dy/dx) So y^3 p'p = 1 p(dp/dy) = y^(-3) Separate, Integrate [tex]\int[/tex]pdp= [tex]\int[/tex]y^-3dy (1/2)p^2 = -(1/2)y^-2 + C p = (2C - (y^-2))^(1/2) 1/p = dx/dy, so dx = (2C - (y^-2))^(1/2) dy And then I would try and integrate that. I can't get the right side (I thought arcsin, but aren't I missing the du (Calc II memory bad), or do I add 0 in the form of + something - something in the numerator? Or is my overall approach off?