Leszek: 4xy ' = x(x+y−1)
2 / : x ≠0
4y ' = (x+y −1)
2 , niech : x+y −1 = u ⇒ 1 + y ' = u ' ⇒ y ' = u ' −1
4 ( du/dx −1) = u
2 ⇒ du/dx = 1 + (u/2)
2
| | du | |
∫ |
| = ∫ dx⇒ 2 arctg(u/2) = x+C |
| | 1 + (u/2)2 | |
czyli : u = 2 tg(x/2 +C)⇒x +y −1 = 2 tg(x/2 +C)⇒ y(x) = 2 tg(x/2 +C) −x+1