齐次微分方程

原式化为：y'=(2yx^2-y^3)/(2x^3)
令u = y/x ,则：y = ux,y'= u + x*du/dx
代入上式可得：
x*du/dx = -1/2*u^3
即：-2du/u^3=dx/x
两边积分,得
1/u^2=lnCx,C为常数
从而：(x/y) = ±(lnCx)^(1/2) (y≠0)
此外,y=0显然也是解.

令：v=y/x,y=xv,dy=vdx+xdv
dy/dx = -(y^2-2xy)/x^2
(vdx+xdv)/dx = 2v - v^2
v+xdv/dx = 2v - v^2
xdv/dx = v - v^2
dv/[v(1 - v)] = dx/x
∫dv/[v(1 - v)] = ∫dx/x
∫dv/v + ∫dv/(1 - v)] = ∫dx/x
lnv - ln(1-v) = lnx
ln(v/(1-v))=lnx+lnc
v/(1-v)=cx
(y/x)/(1-(y/x))=cx
y/(x-y)=cx
y=cx(x-y)