e^y+xy=e……①当x=0时，e^y(0)=e，y(0)=1将①两边对x求导e^y*y'+y+xy'=0……②y'=-y/(e^y+x)当x=0时，y'(0)=-1/e将②两边对x求导e^y*(y')^2+e^y*y''+2y'+xy''=0y''=-[e^y*(y')^2+2y']/(e^y+x)当x=0时，y''(0)=-[e*(-1/e)^2+2*(-1/e)]/e=1/e^2
