x=0
则y+y=e
y=e/2
y+ye^x=e
dy+dye^x=de
dy+e^xdy+y*e^xdx=0
所以y'=dy/dx=-y*e^x/(1+e^x)
所以y'(0)=-(e/2)*1/(1+1)=-e/4
dy=y'dx=-y*e^xdx/(1+e^x)