2xycos〖x2〗−2xy+1dx+sin〖x2〗−x2+3dy=0{2xy cos〖x^2 〗-2xy+1}dx+{sin〖x^2 〗-x^2+3}dy=02xycos〖x2〗−2xy+1dx+sin〖x2〗−x2+3dy=0
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