Question

求解一道微分方程题y'*tany+1/x=e^x*cosy

Answer 1

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)]