设f(x,y)在全平面有连续偏导数,曲线积分∫Lf(x,y)dx+xcosydy在全平面与路径无关,且∫(0,0)(t,t2)f(x,y)dx+xcosydy=t2,求f(x,y).

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问题 设f(x,y)在全平面有连续偏导数,曲线积分∫Lf(x,y)dx+xcosydy在全平面与路径无关,且∫(0,0)(t,t2)f(x,y)dx+xcosydy=t2,求f(x,y).

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答案①∫L(x,y)dx+xcosydy在全平面与路径无关 [*] 积分得f(x,y)=siny+C(x). ②求f(x,y)转化为求C(x). 因f(x,y)dx+xcosydy=sinydx+xcosydy+C(x)dx=sinydx+xdsiny+[*] =d[xsiny+∫0xC(s)ds], 则有 [xsiny+∫0xC(s)ds]|(0,0)(t,t2)=t2, 即tsin t2+∫0tC(s)ds=t2=>sint2+2t2cost2+C(t)=2t,因此 f(x,y)=siny+2x-sinx2-2x2cosx2

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