求一道微分方程,

设u=x y,则dy/dx=du/dx-1代入原方程得du/dx-1=u³ ==>du/dx=u³ 1==>du/(u³ 1)=dx==>1/3[1/(u 1) (2-u)/(u²-u 1)]du=dx==>2/3[2/(u 1)-(2u-1)/(u²-u 1) 3/(u²-u 1)]du=dx==>2/3[2/(u 1)...

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