解：∵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
