求微分方程的通解 {[e^(x+y)]-e^x}dx+{[e^(x+y)]+ey}dy=0 答案是(e^x+1)(e^y+1)=c

问题描述:

求微分方程的通解 {[e^(x+y)]-e^x}dx+{[e^(x+y)]+ey}dy=0 答案是(e^x+1)(e^y+1)=c

[e^(x+y) -e^x]dx +[e^(x+y) +e^y]dy=0
(e^y-1)de^x+(e^x+1)de^y=0
de^x/(e^x+1) +de^y/(e^y-1)=0
dln(e^x+1)+dln(e^y-1)=0
ln(e^x+1)+ln(e^y-1)=C0
(e^x+1)(e^y-1)=C