设[-xy^2dx+x^2ydy]/(x^2+y^2)^m是某二元函数的全微分,则求m
<p>问题：设[-xy^2dx+x^2ydy]/(x^2+y^2)^m是某二元函数的全微分,则求m<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">邓依群的回答： P=-xy^2/(x^2+y^2)^m,Q=x^2y/(x^2+y^2)^m P'=-x/(x^2+y^2)^(2m) =-2xy/(x^2+y^2)^(m+1). Q'=y/(x^2+y^2)^(2m) =2xy/(x^2+y^2)^(m+1). P'=Q',得m=2.
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