Question

(e^(x+y)-e^x)dx+.(e^(x+y)-e^y)dy=0微分方程求通解

Answer 1

(e^(x+y)-e^x)dx+.(e^(x+y)-e^y)dy=0,
(e^x)(e^y-1)dx+(e^y)(e^x-1)dy=0,
(e^x)dx/(e^x-1)=-(e^y)dy/(e^y-1),
[d(e^x-)]/(e^x-1)=- [d(e^y-1)]/(e^y-1),
两边积分，得ln(e^x-1)+C=-ln(e^y-1),即ln(e^x-1)+ln(e^y-1) +C=0,(C为任意常数)
故ln[(e^x-1)(e^y-1)] +C=0,(C为任意常数).
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代入原方程验证，正确。[因(e^x-1).>0,(e^y-1)>0]

Answer 2

分组得：e^(x+y)dx+e^(x+y)dy-e^xdx-e^ydy=0
即：e^(x+y)d(x+y)-e^xdx-e^ydy=0
通解为：e^(x+y)-e^x-e^y=C