（e的x+y次方减去e的X次方）dx+（ex+y次方+e的y次方）dy=0的通解注意有括号..

Question

（e的x+y次方减去e的X次方）dx+（e x+y次方+e的y次方）dy=0的通解 注意有括号..

Answer 1

(e^(x+y)-e^x)dx+(e^(x+y)+e^y)dy=0
(e^xe^y-e^x)dx=-(e^xe^y+e^y)dy
e^x(1-e^y)dx=e^y(e^x+1)dy
e^xdx/(e^x+1)=e^ydy/(1-e^y)
d(e^x+1)/(e^x+1)=-d(1-e^y)/(1-e^y)
ln(e^x+1)=-ln(1-e^y) +C
ln(e^x+1)+ln(1-e^y) =C
ln[(e^x+1)(1-e^y)]=C
(e^x+1)(1-e^y)=C