(2x-y)dx+(2y-x)dy=0的通解
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(2x-y)dx+(2y-x)dy=0的通解
这个是我的解法,应该没有错啊.
死活算不出啦~郁闷死我了!算了很久!
答案是 x^2-xy+y^2=C
![](http://img.wesiedu.com/upload/b/dd/bdd187e03ff3cdbdd022074de1fe904a.jpg)
死活算不出啦~郁闷死我了!算了很久!
答案是 x^2-xy+y^2=C
![(2x-y)dx+(2y-x)dy=0的通解](/uploads/image/z/18457935-15-5.jpg?t=%282x-y%29dx%2B%282y-x%29dy%3D0%E7%9A%84%E9%80%9A%E8%A7%A3)
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
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