兩邊對(duì)x求導(dǎo)：y'=e^y+xe^y* y'得：y'=e^y/(1-xe^y)=(y-1)/x/(1-y+1)=(y-1)/[x(2-y)]y"=[y'x(2-y)-(y-1)(2-y-xy')]/[x^2(2-y)^2]=[(y-1)-(y-1)(2-y-(y-1)/(2-y)]/[x^2(2-y)^2]=(y-1)^2(3-y)/[x^2(2-y)^3]