设∫(0,0)(t,t2)f(x,y)dx+xcosydy=t2,f(x,y)有一阶连续偏导数,求f(x,y).

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问题 设∫(0,0)(t,t2)f(x,y)dx+xcosydy=t2,f(x,y)有一阶连续偏导数,求f(x,y).

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答案因为曲线积分与路径无关,所以有cosy=f′y(x,y),则f(x,y)=siny+C(x),而∫(0,0)(t,t2)f(x,y)dx+xcosydy=t2,即∫0tC(x)dx+∫0t2tcosydy=t2,两边对t求导数得C(t)=2t-sint2-2t2cost2,于是f(x,y)=siny+2x-sinx2-2x2cosx2

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