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

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