求微分方程y'+y/x=cosx/x满足条件x=π时y=1的特解

网友回答

∵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
======以下答案可供参考======
供参考答案1:
令t[x] = y[x]·x
t'[x] = y[x] + y'[x]·x
所以t'[x]/x = y[x]/x + y'[x] = Cos[x]/x
t'[x] = Cos[x]
t[x] = Sin[x] + C
y[x] = t[x]/x = (Sin[x] + C)/x
y[π] = 1
所以C = πy[x] = (Sin[x] + π)/x