y'+y=1
dy/dx+y=1
dy/(1-y)=dx
-d(1-y)/(1-y)=dx
两边同时积分得
-ln(1-y)=x+lnC
ln[C(1-y)]=-x
e^(-x)=C(1-y)
y=1-1/(Ce^x)
所以
y=1-C/e^x