设u=x+y，则y=f（u）∴dydx＝f′(u)dudx＝f′(u)(1+dydx)解得：dydx＝f′(u)1−f′(u)∴d2ydx2＝ddx(f′(u)1−f′(u))＝ddu(f′(u)1−f′(u))•dudx=f″(u)[1−f′(u)]+f′(u)f″(u)[1−f′(u)]2•(1+f′(u)1−f′(u...