Question

[e^(x+y)-e^x]dx+[e^(x+y)+e^y]dy=0,求i通解

Answer 1

解：∵[e^(x+y)-e^x]dx+[e^(x+y)+e^y]dy=0
==>(e^y-1)e^xdx+(e^x+1)e^ydy=0
==>e^xdx/(e^x+1)+e^ydy/(e^y-1)=0
==>d(e^x)/(e^x+1)+d(e^y)/(e^y-1)=0
==>ln(e^x+1)+ln│e^y-1│=ln│C│ (C是积分常数)
==>(e^x+1)(e^y-1)=C
∴原微分方程的通解是(e^x+1)(e^y-1)=C (C是积分常数)。