化dy/dx=(x+3y-5)/(2x-y+1)为课分离变量方程

解方程x+3y-5=0,2x-y+1=0
得x=2/7,y=11/7
令u=[y- (11/7)]/[x-(2/7)]
则y-(11/7)=[x-(2/7)]u
所以dy/dx=d(y-(11/y))/d(x-(2/7)) = [x-(2/7)]u'+u
另外(x+3y-5)/(2x-y+1)
上下同除x-2/7
=(1+3u)/(2-u)
所以
[x-(2/7)]u'+u=(1+3u)/(2-u)
所以du/dx = {[(1+3u)/(2-u)]-u} / [x-(2/7)]
所以du/{[(1+3u)/(2-u)]-u} = dx/[x-(2/7)]
已化为分离变量方程