解微分方程dy/dx=((x+y-1)^2)/((x+y+1)^2)

两边加1得
d(x+y)/dx = [(x+y-1)/(x+y+1)]^2 + 1
设x+y = u
那么du/dx = [(u-1)^2+(u+1)^2]/(u+1)^2 = 2(u^2+1)/(u+1)^2
所以(u+1)^2/(u^2+1) du = 2dx
积分得
u + Ln[2 - 2 (1 + u) + (1 + u)^2] = 2x + C
所以方程的通解为
x+y + Ln[2 - 2(x+y+1) + (x+y+1)^2 ] = 2x+c