①dy/dx+xy=xy²dy=x(y²-y)dxxdx=dy/(y²-y)=dy/(y-1)-dy/y两边分别积分,得x²/2+C=ln[(y-1)/y]整理得y=1/[1-Ce^(x²/2)]②y'-y=x齐次方程y'-y=0的通解为y=Ce^x设非齐次y'-y=x的通解为y=ue^x,...