y'^3 -4xyy'+8y^2 =0.

skipjack Forum Staff Dec 2006 21,482 2,472 Jan 18, 2020 #2 The general solution is probably very complicated, but there are some solutions that are easy to find: $y = 0$, $y = 4x^3/27$, and (for real x) two piecewise-defined solutions based on those. Reactions: topsquark and idontknow

The general solution is probably very complicated, but there are some solutions that are easy to find: $y = 0$, $y = 4x^3/27$, and (for real x) two piecewise-defined solutions based on those.