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Differential Calculus> differentiation...

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if y= (3/x^2 - 1/b^2)sin(x/b + A) - 3/bx*cos(x/b+A) satisfies the relationd^2y/dx^2 + (1/b^2-k/x^2)y=0then find the value of k

alisha gupta , 11 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans: 6

$y = (\frac{3}{x^{2}}-\frac{1}{b^{2}})sin(\frac{x}{b}+A)-\frac{3}{bx}cos(\frac{x}{b}+A)$ $y^{'} = (\frac{3}{x^{2}}-\frac{1}{b^{2}})cos(\frac{x}{b}+A).\frac{1}{b}+sin(\frac{x}{b}+A).\frac{-6}{x^{3}}-[\frac{3}{bx}sin(\frac{x}{b}+A).\frac{-1}{b}-\frac{3}{bx^{2}}cos(\frac{x}{b}+A)$

$y^{'} = cos(\frac{x}{b}+A)[\frac{6}{bx^{2}}-\frac{1}{b^{3}}]+sin(\frac{x}{b}+A)[\frac{3}{b^{2}x}-\frac{6}{x^{3}}]$

$y^{''} = \frac{3}{bx}cos(\frac{x}{b}+A)[\frac{1}{b^{2}}-\frac{6}{x^{2}}]+sin(\frac{x}{b}+A)[\frac{1}{b^{4}}-\frac{9}{b^{2}x^{2}}+\frac{18}{x^{4}}]$

$y^{''}+(\frac{1}{b^{2}}-\frac{6}{x^{2}})y = 0$

k = 6

Thanks & Regards

Jitender Singh

IIT Delhi

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