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通解
问题描述:
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通解
答
e^(x+y)-e^x+e^y=c
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通解
e^(x+y)-e^x+e^y=c