两边对x求导：
y'e^y+(1+y')cos(x+y)=0,1）
这里可得到y'=-cos(x+y)/[e^y+cos(x+y)]
再对1）求导：
y"e^y+(y')^2e^y+y"cos(x+y)-(1+y')^2 sin(x+y)=0
代入y' 得;
y"(e^y+cos(x+y)]+[cos(x+y)]^2/[e^y+cos(x+y)]^2* e^y-(1-cos(x+y)/(e^y+cos(x+y)))^2sin(x+y)=0
得:y"=-[cos(x+y)]^2/[e^y+cos(x+y)]^2* e^y+(1-cos(x+y)/(e^y+cos(x+y)))^2sin(x+y)}/(e^y+cos(x+y))