du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)

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du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)

du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)
du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)

du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)
因为du/dx=2xycos(x^2y)
所以u(x,y)=∫2xycos(x^2y)dx
=∫cos(x^2y)d(x^2y)
=sin(x^2y)+A(y) 其中A(y)是关于y的任意函数
因为du/dy=x^2cos(x^2y)
所以x^2cos(x^2y)+A'(y)=x^2cos(x^2y)
所以A'(y)=0
A(y)=C
所以u(x,y)=sin(x^2y)+C 其中C是任意常数