y'=(x-y+1)/(x+y-3),
令x+y-3=t,
则y=t-x+3,y'=t'-1,
所以t'-1=(2x-t-2)/t,
t'=2(x-1)/t,分离变量法求
t=正负根号((x-1)^2/2+c),
y=正负根号((x-1)^2/2+c)-x+3