If $y = 3 cos (log x) + 4 sin (log x)$ , show that $x^{2} y_{2} + x y_{1} + y = 0$

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Solution

$y = 3 cos (log x) + 4 sin (log x) y_{1} = \frac{d y}{d x} = 3 (- sin (log x)) \times \frac{1}{x} + 4 (cos (log x)) \times \frac{1}{x}$
$y_{2} = 3 (- cos (log x)) \times \frac{1}{x^{2}} + 3 (sin (log x)) \times \frac{1}{x^{2}} + 4 (- sin (log x)) \times \frac{1}{x^{2}} - 4 (cos (log x)) \times \frac{1}{x^{2}}$ ...By Using Product rule
Now, by putting the value of $y_{2}$ & $y_{1}$
$x^{2} y_{2} + x y_{1} + y = 3 (- cos (log x)) + 3 (sin (log x)) - 4 (sin (log x)) - 4 (cos (log x)) - 3 (sin (log x)) + 4 (cos (log x)) + 3 cos (log x) + 4 sin (log x)$
$= 0$

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