设y(x)满足微分方程(e^x)yy'=1,且y(0)=1,则y=

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设y(x)满足微分方程(e^x)yy'=1,且y(0)=1,则y=
设y(x)满足微分方程(e^x)yy'=1,且y(0)=1,则y=

设y(x)满足微分方程(e^x)yy'=1,且y(0)=1,则y=
(e^x)yy'=1所以
ydy=e^-xdx 两边积分得
1/2y^2=-e^-x+c
又因为y(0)=1 代入解得
c=3/2
所以y^2=-2e^-x+3

y=1/e^x

yy'=e^(-x)
y(dy/dx)=e^(-x)
ydy=e^(-x)dx
1/2y^2=-e^(-x)
y^2= -2e^(-x)