设Q(x,y)在平面xOy上具有一阶连续的偏导数,且∫L2xydx+Q(x,y)dy与路径无关,且对任意的t有∫(0,0)(t,1)2xydx+Q(x,y)dy=∫(0,0)(1,t)2xydx+Q(x,y)dy,求Q(x,y).

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问题 设Q(x,y)在平面xOy上具有一阶连续的偏导数,且∫L2xydx+Q(x,y)dy与路径无关,且对任意的t有∫(0,0)(t,1)2xydx+Q(x,y)dy=∫(0,0)(1,t)2xydx+Q(x,y)dy,求Q(x,y).

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答案因为曲线积分与路径无关,所以[*]=2x,于是Q(x,y)=x2+φ(y). 由∫(0,0)(t,1)2xydx+Q(x,y)dy=∫(0,0)(1,t)2xydx+Q(x,y)dy,得 t2+∫01φ(y)dy=t+∫0tφ(y)dy,两边对t求导数得1+φ(t)=2t,φ(t)=2t-1, 所以Q(x,y)=x2+2y一1.

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