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

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