(x-2xy-y²)y' + y² = 0
(x-2xy-y²)dy + y²dx = 0
xdy - xd(y²) - y²dy + y²dx = 0
(x-y²)dy = xd(y²) - y²dx = x²d(y²/x)
(1 - y²/x)dy = xd(y²/x)
令y²/x = u,即x = y²/u
代入得
(1-u)dy = y²du/u
即dy/y² = du/[u(1-u)] = du/u + du/(1-u)
积分得
-1/y = ln|u/(1-u)| + C
|u/(1-u)| = Ce^(-1/y)
u = 1/[1±Ce^(1/y)]
x = y²[1+Ce^(1/y)]（C是任意常数,加或减一个任意常数都可用+C表示）
这是通解