t(e^y)+y+1=0求导,得到e^y+te^yy'+y'=0得到y'=-e^y/(te^y+1)x+t(1-t)=0求导,得到x'=2t-1那么dy=[-e^y/(te^y+1)]dx/(2t-1)=-e^ydx/(te^y+1)(2t-1)t=0时,x=0,y=-1因此dy|t=0 =(1/e)dx