Question

常微分[e（x+y）的次方-e的x次方]dx+{e(x+y)的次方+e的y次}dy=0的通解

Answer 1

移项 [exp（x+y）-exp(x)]dx = -[exp(x+y)+exp(y)]dy
化简得 {exp(x)/[1+exp(x)]}dx = {exp(y)/[1-exp(y)]}dy
积分得 ln[1+exp(x)] + C = -ln[1-exp(y)]
进一步化简得 C*[1+exp(x)][1-exp(y)] = 1