∵y'+y/x=cosx/x==>xy'+y=cosx
==>xdy+ydx=cosxdx
==>d(xy)=d(sinx)
∴xy=sinx+C (C是积分常数)
∵微分方程满足条件x=π时y=1
∴π*1=sinπ+C==>C=π
故原方程的解是：xy=sinx+π