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(1+2y)xdx+(1+x^2)dy=0的微分方程的通解.答案是(1+x^2)(1+Y^2)=c.
人气:136 ℃ 时间:2020-06-04 03:03:54
解答
  (1+2y)xdx+(1+x^2)dy=0  分离变量:  x/(1+x^2) dx +1/(1+2y) dy=0  积分:  ∫x/(1+x^2)dx+∫1/(1+2y)dy=0  1/2∫2x/(1+x^2)dx+1/2∫1/(1+2y)d2y=0  1/2∫1/(1+x^2)dx^2+1/2∫1/(1+2y)d2y=0  ...厉害谢谢啊
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