(cosy+ycosx)dx+(sinx-siny)dy=0
人气：403 ℃ 时间：2020-01-14 17:36:46
解答
(cosxsiny)dx+(sinxcosy)dy=0
sinydsinx+sinxdsiny=0
dsinx/sinx+dsiny/siny=0
d(lnsinx)+d(lnsiny)=0
d(ln(sinxsiny))=0
ln(sinxsiny)=C1
sinxsiny=C2
其中C1和C2为任意常数亲，答案绝对正确，求好评
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