∵dy/dx=1/(x–y^2)==>dx-xdy+y^2dy=0==>e^(-y)dx-xe^(-y)dy+y^2e^(-y)dy=0 (等式两端同乘e^(-y))==>d(xe^(-y))-d((y^2+2y+2)e^(-y))=0 ==>xe^(-y)-(y^2+2y+2)e^(-y)=C (C是常数)==>x=y^2+2y+2+Ce^y∴原方程的通解是...