If $y = a cos (log x) + b sin (log x),$ Find the value of $x^{2} y_{2} + x y_{1} + y$ .

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Solution

Given $y = a c o s (l o g x) + b s i n (l o g x)$

$⟹ y_{1} = - a s i n (l o g x) (\frac{1}{x}) + b c o s (l o g x) (\frac{1}{x})$

$⟹ y_{1} = (- a s i n (l o g x) + b c o s (l o g x)) (\frac{1}{x})$ ------(1)

$⟹ y_{2} = (- a s i n (l o g x) + b c o s (l o g x)) (- \frac{1}{x^{2}}) + (\frac{1}{x}) (- \frac{a}{x} c o s (l o g x) - \frac{b}{x} s i n (l o g x))$

$⟹ y_{2} = (a s i n (l o g x) - b c o s (l o g x) - a c o s (l o g x) - b s i n (l o g x)) (\frac{1}{x^{2}})$ -----(2)

Therefore, $x^{2} y_{2} + x y_{1} + y$ $= a s i n (l o g x) - b c o s (l o g x) - a c o s (l o g x) - b s i n (l o g x) - a s i n (l o g x) + b c o s (l o g x) + a c o s (l o g x) + b s i n (l o g x)$

$⟹ x^{2} y_{2} + x y_{1} + y = 0$

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