(x+y)dx+(3x+3y-4)dy=0
问题描述：
(x+y)dx+(3x+3y-4)dy=0
答
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...