来自汤伟平的问题
(x+y)dx+(3x+3y-4)dy=0
(x+y)dx+(3x+3y-4)dy=0
1回答
2020-08-06 21:11
(x+y)dx+(3x+3y-4)dy=0
(x+y)dx+(3x+3y-4)dy=0
令u=x+y,则du/dx=1+dy/dx,既dy/dx=du/dx-1.原方程可变为dy/dx=-(x+y)/(3x+3y-4),既du/dx-1=-u/(3u-4).这已经是可分离变量的微分方程.(3u-4)/(2u-4)du=dx,两边积分,并化简的(u带回x+y),x+3y+ln(x+y-2)^2...