The correct option is B 1y
(x+2y3)dydx=y, this equation involves the term y3 is not linear, if we take y as the dependent, so we can write the equation as ydxdy=x+2y3
⇒ ydxdy−x=2y3, which is linear differential equation in which x is as the dependent variable
dxdy−xy=2y2
∴ I.F. =e∫−1ydy=e−logy=elog1y=1y