dy/dx=2xy(y-1)dy/[y(y-1)]=2xdx等式两边同时积分∫dy/[y(y-1)]=∫2xdx∫[1/(y-1) -1/y]dyln[(y-1)/y]=x²+C(y-1)/y=e^(x²+C)y=1/[1-e^(x²+C)]x=0，y=-1代入，得C=ln2y=1/[1-e^(x²+C)]=1/[1-2e^(x²)]所求特解为：y=1/[1-2e^(x²)]
