求微分方程的通解.[1+2e^(x/y)]dx+ 2e^(x/y)*[1-x/y]dy=0.

令x/y=p
x=py
x'=p+p'y
[1+2e^(x/y)]dx+ 2e^(x/y)*[1-x/y]dy=0
[1+2e^(x/y)]dx/dy+ 2e^(x/y)*[1-x/y]=0
(1+2e^p)(p+p'y)+2e^p*(1-p)=0
p+p'y=-2e^p*(1-p)/(1+2e^p)
p'y=-2e^p*(1-p)/(1+2e^p)-p=(-2e^p+2e^p*p-p-2e^p*p)/(1+2e^p)
=(-2e^p-p)/(1+2e^p)
(1+2e^p)/(2e^p+p)dp=-dy/y
d(p+2e^p)/(2e^p+p)=-dy/y
ln(2e^p+p)=-lny+C1
2e^p+p=C/y
反代即可