一道 微分方程的题 , dy/dx=3x-y/x+y 求方程的解,求详解,
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dy/dx= (3x-y)/(x+y)= (3-y/x)/(1+y/x)
y/x=u y=ux y'=u+xu'
u+xu'= (3-u)/(1+u)
xu'= (3-u)/(1+u)-u= (3+u)(1-u)/(1+u)
(1+u)du/(3+u)(1-u)=dx/x
(1/2)[1/(3+u)+1/(1-u)]du=dx/x
ln(3+u)-ln(1-u)=2lnx+lnC
(3+u)/(1-u)=Cx^2
(3+y/x)/(1-y/x)=Cx^2
或：
(3x+y)/(x-y)=Cx^4
y/x=u y=ux y'=u+xu'
u+xu'= (3-u)/(1+u)
xu'= (3-u)/(1+u)-u= (3+u)(1-u)/(1+u)
(1+u)du/(3+u)(1-u)=dx/x
(1/2)[1/(3+u)+1/(1-u)]du=dx/x
ln(3+u)-ln(1-u)=2lnx+lnC
(3+u)/(1-u)=Cx^2
(3+y/x)/(1-y/x)=Cx^2
或：
(3x+y)/(x-y)=Cx^4
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