(e^x+y - e^x)dx+(e^x+y + e^y)dy=0
化简
(e^y-1)e^xdx+e^y(e^x+1)dy=0
即(e^y-1)d(e^x)+(e^x+1)d(e^y)=0
设e^x=p e^y=t
则(t-1)dp=-(p+1)dt
d(p+1)/(p+1)=-d(t-1)/(t-1)
Ln|p+1|=-Ln|t-1|+C
即(p+1)(t-1)=C
即(e^x+1)(e^y-1)=C