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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
答案：↓↓↓ 邓依群的回答： P=-xy^2/(x^2+y^2)^m,Q=x^2y/(x^2+y^2)^m P'=-x[2y(x^2+y^2)^m-2my^3(x^2+y^2)^(m-1)]/(x^2+y^2)^(2m) =-2xy[x^2+(1-m)y^2]/(x^2+y^2)^(m+1). Q'=y[2x(x^2+y^2)^m-2mx^3(x^2+y^2)^(m-1)]/(x^2+y^2)^(2m) =2xy[y^2+(1-m)x^2]/(x^2+y^2)^(m+1). P'=Q',得m=2. | 
