求解一道微分方程题y'*tany+1/x=e^x*cosy

y'tany+1/x=e^xcosy
u=1/cosy,u'=y'siny/cos^2y=y'tany/cosy,y'tany=u'cosy=u'/u
u'/u+1/x=e^x/u
u'+u/x=e^x
xu'+u=xe^x
(xu)'=xe^x=(xe^x)'-e^x=[(x-1)e^x]'
xu=(x-1)e^x,{要不要+C?}
u=(x-1)e^x/x=1/cosy
cosy=e^(-x)x/(x-1)
y=arccos[e^(-x)x/(x-1)]