求(y3-3xy2-3x2y)dx+(3xy2-3x2y-x3+y2)dy=0的通解.

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问题 求(y3-3xy2-3x2y)dx+(3xy2-3x2y-x3+y2)dy=0的通解.

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答案将原给方程通过视察分项组合. (y3-3xy2-3x2y)dx+(3xy2-3x2y-x3+y2)dy =(y3dx+3xy2dy)-3xy(ydx+xdy)-(3x2ydx+x3dy)+y2dy =0, 即 d(xy3)-[*]d(xy)2-d(x3y)+[*]d(y3)=0, d[xy3-[*](xy)2-x3y+[*]y3]=0, 所以通解为xy3-[*]x2y2-x3y+[*]y3=C,其中C为任意常数.

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