(2x-y)dx+(2y-x)dy=0的通解

1个回答

  • 2dx/x=(1-2u)/(1-u+u^2)du

    2dx/x=(1-2u)/[(u-1/2)^2+3/4] du

    2dx/x=- d(u-1/2)^2/[(u-1/2)^2+3/4],这一步关键是将分子看成是(u-1/2)^2的微分.

    积分:

    2ln|x|+c1=-ln[(u-1/2)^2+3/4]

    C/x^2=(u-1/2)^2+3/4

    C/x^2=y^2/x^2-y/x+1

    C=y^2-xy+x^2