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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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Verified Answer
$\frac{d y}{d x}=y_1=\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x}$
$\Rightarrow \quad x_1=-3 \sin (\log x)+4 \cos (\log x)$
$x_2+y_1=-\frac{3 \cos (\log x)}{x}-\frac{4 \sin (\log x)}{x}=-\frac{y}{x}$
$\Rightarrow x^2 y_2+x y_1+y=0 . \quad$ Hence proved.
$\Rightarrow \quad x_1=-3 \sin (\log x)+4 \cos (\log x)$
$x_2+y_1=-\frac{3 \cos (\log x)}{x}-\frac{4 \sin (\log x)}{x}=-\frac{y}{x}$
$\Rightarrow x^2 y_2+x y_1+y=0 . \quad$ Hence proved.
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