∵x(y^2+2xyy')=cos(xy^2)==>xd(xy^2)/dx=cos(xy^2)==>d(xy^2)/cos(xy^2)=dx/x==>ln│(1+sin(xy^2))/cos(xy^2)│=ln│x│+ln│C│ (C是常数)==>(1+sin(xy^2))/cos(xy^2)=Cx==>1+sin(xy^2)=Cxcos(xy^2)∴原方程的通解...