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方程y'^3+2xy'-y=0的通解
人气:340 ℃ 时间:2020-03-27 04:33:49
解答
∵令y'=p,则代入原方程得y=p³+2xp.(1)
==>p=3p²p'+2p+2xp'(两端对x求导数)
==>3p²dp+2xdp+pdx=0
==>3p³dp+2xpdp+p²dx=0(两端同乘p)
==>3d(p^4)+4d(xp²)=0
==>3p^4+4xp²=C(C是积分常数)
==>x=C/p²-3p²
==>y=2C/p-5p³
∴原方程的通解参数形式是x=C/p²-3p²,y=2C/p-5p³(C是积分常数,p是参数).
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