求（y+x^3)dx=2xdy的通解

急，谢谢！

设u=y+x^3
有u'=y'+3x^2, 因此y'=u'-3x^2
代入方程u-2xu'+6x^3=0
再设v=u/(x^(1/2))
有v'=u'/(x^(1/2))-u/(2x^(3/2))=(2xu'-u)/(2x^(3/2))=3x^(3/2)
解得v=6/5*x^(5/2)+C
故u=6/5*x^3+Cx^(1/2)
最后y=1/5*x^3+Cx^(1/2)

e^(-x)dx=xdy ==>e^(-x)]dx-xdy=0 ==>ydx+x²x=c-e^(-x) (c是积分常数) ==>y=x[c-e^(-x)] ∴原方程的通解是y=x[c-e^(-x)] (c是积分常数);(xdy-ydx)/：∵[y+x²e^(-x)dx ==>x²x)=e^(-x)dx ==>y/=e^(-x)dx ==>xdy-ydx=x²解;d(y/