Question Number 82579 by M±th+et£s last updated on 22/Feb/20
$${if}\:{y}={cos}\left({ln}\left({x}\right)\right)+{sin}\left({ln}\left({x}\right)\right) \\ $$$${show}\:{that} \\ $$$${y}''+{y}^{'} +{y}=\mathrm{0} \\ $$
Answered by mind is power last updated on 23/Feb/20
$${y}'=\frac{\mathrm{1}}{{x}}\left(−{sin}\left({ln}\left({x}\right)\right)+{cos}\left({ln}\left({x}\right)\right)\right. \\ $$$${y}''=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(−\mathrm{2}{cos}\left({ln}\left({x}\right)\right)\right. \\ $$$${x}^{\mathrm{2}} {y}''+{xy}'+{y}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$