Question

1+y＇=e的y次方,求微分方程的通解!

Answer 1

1+dy/dx=e^y

dx+dy=e^y*dx

dy/(e^y-1)=dx

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

左边对y进行积分,右边对x进行积分,得

-y+ln(e^y-1)=x+c