（e的x+y次方-e的x次方）dx+(e的x+y次方+e的y次方）dy=0求通解能有过程

网友回答

e^(x + y) - e^x + [e^(x + y) + e^y] • dy/dx = 0
[e^(x + y) + e^y] • dy/dx = e^x - e^(x + y) = e^x • (1 - e^y)
dy/dx = (e^x)/(e^y) • (1 - e^y)/(e^x + 1)
∫ (e^y)/(1 - e^y) dy = ∫ (e^x)/(e^x + 1) dx
∫ - d(1 - e^y)/(1 - e^y) = ∫ d(e^x + 1)/(e^x + 1)
ln|1 - e^y| = - ln(e^x + 1) + lnC
1 - e^y = C/(e^x + 1)
e^y = 1 - C/(e^x + 1)
y = ln[1 - C/(e^x + 1)]