y'=(x+y)^2 若要用變數分離的方法的話 要怎麼解啊？

令 z = x + y
dz = dx + dy

dy/dx = (x+y)^2 = z^2
dy = z^2 dx
dz - dx = z^2 dx
dz = (1+z^2) dx
dz/(1+z^2) = dx
∫ [ 1/(1+z^2) ] dz = ∫ dx + C
arctan Z = x + C , 請參考底下註解
arctan(x+y) = x + C
tan(x+C) = x+y
y = - x + tan(x+C)

Ans: y = - x + tan(x+C)

驗算:
x + y = tan(x+C)
dy/dx = -1 + sec^2 (x+C) = tan^2 (x+C) = (x+y)^2
故驗算無誤

註:
∫ [ 1/(1+z^2) ] dz = arctan Z + C
pf :
令 z = tanθ , 則 dz = sec^2 θ dθ , 1+z^2 = sec^2 θ
∫ [ 1/(1+z^2) ] dz
= ∫ ( 1 / sec^2 θ ) sec^2 θ dθ
= ∫ dθ
= θ + C
= arctan z + C