If y = 3 cos (log x) + 4 sin (log x), show that $x^{2} y_{2} + x y_{1} + y = 0$
where $y_{1}$ & $y_{2}$ are the successive derivatives of y.

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Solution

$y = 3 c o s (l o g x) + 4 s i n (l o g x) y_{1} = \frac{- 3 s i n (l o g x)}{x} + \frac{4 c o s (l o g x)}{x} \Rightarrow x y_{1} = - 3 s i n (l o g x) + 4 c o s (l o g x) y_{2} = \frac{d y_{1}}{d x} = \frac{x (\frac{- 3 c o s (l o g x)}{x} - \frac{4 s i n (l o g x)}{x}) - 1. (\frac{- 3 s i n (l o g x)}{x} + \frac{4 c o s (l o g x)}{x})}{x^{2}} \Rightarrow y_{2} = \frac{- (3 c o s (l o g x) + 4 s i n (l o g x)) - (\frac{- 3 s i n (l o g x) + 4 c o s (l o g x)}{x})}{x^{2}} \Rightarrow x^{2} y_{2} = - (y) - (y_{1}) \Rightarrow x^{2} y_{2} + y_{1} + y = 0$

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