求微分方程 dy/dx=(x+y)^2

变换u＝x＋y，则y'＝u'－1，方程化为u'－1＝u^2，分离变量：du/(1+u^2)＝dx，两边积分：arctanu＝x＋C，所以u＝tan(x+C)，所以y＝tan(x+C)－x

变换u＝x＋y，v=x-y,则
x=(u+v)/2,y=(u-v)/2
dx=(du+dv)/2,dy=(du-dv)/2
dy/dx=(du-dv)/(du+dv)=u^2
整理得dv/du=(1-u^2)/(1+u^2)=2/(1+u^2)-1
v=2arctan(1+u^2)-u+C
剩下自己解吧

印错了吧，dy/dx = (x-y)²的通解才是1/2ln[(1+x-y)/(1-x+y)]=x+C
令x-y = u
则du = dx-dy
dy = dx-du
dy/dx = 1 - du/dx = (x-y)² = u²
du/dx = 1 - u²
dx = du/(1-u²) = du/2(1-u) + du/2(1+u)
两边积分得
x + C = 1/2 * [-ln(1-u)] + 1/2 * ln(1+u)
即1/2 * ln[(1+u)/(1-u)] = x + C
把u = x-y代入得通解为
1/2 * ln[(1+x-y)/(1-x+y)] = x + C