dy/dx=e^(x+y)微分方程的通解

Question

dy/dx=e^(x+y)微分方程的通解

Answer 1

令u=x+y
u'=1+y'
y'=e^u 化为：
u'-1=e^u
因此有：du/dx=e^u+1
du/(e^u+1)=dx
d(e^u)/[1/e^u-1/(e^u+1)]=dx
ln(e^u)-ln(e^u+1)=x+c1
e^u/(e^u+1)=ce^x
e^(x+y)/[e^(x+y)+1]=ce^x