dy/dx=1/(x-y^2) 求微分方程的通解

解法一：∵dy/dx=1/(x-y^2)==>dx-(x-y^2)dy=0==>e^(-y)dx-xe^(-y)dy=-y^2e^(-y)dy (等式两端同乘e^(-y))==>d(xe^(-y))=d((y^2+2y+2)e^(-y))==>xe^(-y)=(y^2+2y+2)e^(-y)+C (C是积分常数)==>x=y^2+2y+2+Ce^y∴原方程...