The solution of differential equation dydx=1−x(y−x)−x3(y−x)3 is
y−x−2=cex2−1-x2
y+x−2=cex2−x2
y−x−2=cex2−1+x2
None of these
Put y−x=v⇒dvdx+Vx=−x3v2 v−3dvdx+xv−2=−x3 put V−2=u⇒dudx−2xu=2x3 G.S. is (y−x)−2=cex2−1+x2