请教一道微分方程（x-2xy-y^2)y'+y^2=0求通解.
3个回答

(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

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