Question

微分方程y'+y=1的通解.

Answer 1

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