设u=√(y/x)u'x=(-1/2)x^(-3/2)y^(1/2)u'y=(1/2)(xy)^(-1/2)那么原式变成了arctanu=(1/u^2)所以(u^2)arctanu=1两边取全微分得到[2uarctanu+u^2/(1+u^2)]*[(u'x)dx+(u'y)dy]=0那么(u'x)dx+(u'y)dy=0dy=-(y/x)dx...