Question

微分方程(1+x)y'+1=2e^(-y)的通解为__________

Answer 1

(1+x)y'+1=2e^(-y)

(1+x)y'=2e^(-y)-1

dy/[2e^(-y)-1]=dx/(1+x)

e^ydy/[2-e^y]=dx/(1+x)

积分得通-ln(2-e^y)+lnC=ln(1+x)

或：（1+x)(2-e^y)=C