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2xy Cosx

因为(x^2+1)y'+2xy=[(x^2+1)y]'=cosx 所以(x^2+1)y=c+sinx y=(sinx+c)/(x^2+1),c为任意常数

你好,满意请采纳哦! P(x,y)=2xy^3-y^2cosx,Q(x,y)=1-2ysinx+3x^2y^2 αP/αy=αQ/αx=6xy^2-2ycosx 因此本题积分与路径无关,可自选积分路线

因为(x^2+1)y'+2xy=[(x^2+1)y]'=cosx 所以(x^2+1)y=c+sinx y=(sinx+c)/(x^2+1),c为任意常数

P(x,y)=2xy^3-y^2cosx,Q(x,y)=1-2ysinx+3x^2y^2 αP/αy=αQ/αx=6xy^2-2ycosx,所以曲线积分与路径无关,替换积分路径为从〇(0,0)...

如图

若原函数为f(x,y)=x+x2y3-y2sinx+y+c,则分别对x,y求偏导时,df(x,y)=(1+2xy3-y2cosx)dx+(1-2ysinx+3x2y2)dy,所以A选项正确.而f(x,y)=y-y2sinx+x2y3+c,df(x,y)=(-y2cosx+2xy3)dx+(1-2ysinx+3x2y2)dy,所以选项B错误.而...

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