求微分方程y'-1/x=xsinx的通解

Question

如题所述

Answer 1

y'=dy/dx=xsinx+1/x
所以dy=(xsinx+1/x)dx
两边积分
y=-∫xdcosx+ln|x|
=-xcosx+∫cosxdx+ln|x|
=-xcosx+sinx+ln|x|+C

Answer 2

y'-1/x=xsinx
y' = xsinx +1/x

y
= ∫ (xsinx +1/x) dx
= ∫ xsinx dx +ln|x|
= -∫ xdcosx +ln|x|
= -xcosx + ∫ cosx dx+ln|x|
= -xcosx + sinx +ln|x| + C本回答被提问者采纳