Question

设函数y=y(x)是由方程y+ye^x=e确定的隐函数'求y'(0)和dy

Answer 1

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)