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Question Number 43489 by peter frank last updated on 11/Sep/18

if y=cos(logx) +3sin(logx) prove that x^2 (d^2 x/dx^2 )+x(dy/dx)+y=0

$${if}\:{y}={cos}\left({logx}\right)\:+\mathrm{3}{sin}\left({logx}\right)\:{prove}\:{that} \\ $$$${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {x}}{{dx}^{\mathrm{2}} }+{x}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

(dy/dx)=−sin(logx)×(1/x)+3cos(logx)×(1/x) x(dy/dx)=−sin(logx)+3cos(logx)= x(d^2 y/dx^2 )+(dy/dx)=−cos(logx)×(1/x)−3sin(logx)×(1/x) x^2 (d^2 y/dx^2 )+x(dy/dx)=−y x^2 (d^2 y/dx^2 )+x(dy/dx)+y=0

$$\frac{{dy}}{{dx}}=−{sin}\left({logx}\right)×\frac{\mathrm{1}}{{x}}+\mathrm{3}{cos}\left({logx}\right)×\frac{\mathrm{1}}{{x}} \\ $$$${x}\frac{{dy}}{{dx}}=−{sin}\left({logx}\right)+\mathrm{3}{cos}\left({logx}\right)= \\ $$$${x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{{dy}}{{dx}}=−{cos}\left({logx}\right)×\frac{\mathrm{1}}{{x}}−\mathrm{3}{sin}\left({logx}\right)×\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}\frac{{dy}}{{dx}}=−{y} \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$

Commented by peter frank last updated on 11/Sep/18

much respect sir

$${much}\:{respect}\:{sir} \\ $$

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