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推荐于2019-06-06 · 知道合伙人教育行家
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du+dv=dx+dy,
sinudy+ycosudu=sinvdx+xcosvdv,即ycosudu-xcosvdv=sinvdx-sinudy.
看作是以du,dv为未知量的二元一次方程组,解得
du=((xcosv+sinv)dx+(xcosv-sinu)dy)/(ycosu+xcosv),
dv=((ycosu-sinv)dx+(sinu+ycosu)dy)/(ycosu+xcosv).
sinudy+ycosudu=sinvdx+xcosvdv,即ycosudu-xcosvdv=sinvdx-sinudy.
看作是以du,dv为未知量的二元一次方程组,解得
du=((xcosv+sinv)dx+(xcosv-sinu)dy)/(ycosu+xcosv),
dv=((ycosu-sinv)dx+(sinu+ycosu)dy)/(ycosu+xcosv).
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