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If $y=a \cos (\log x)+b \sin (\log x)$, where $a, b$ are parameters, then $x^2 y^{\prime \prime}+x y^{\prime}$ is equal to
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The correct answer is:
$y$
We have,
$$
\begin{aligned}
& y=a \cos (\log x)+b \sin (\log x) \\
& y^{\prime}=\frac{-a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x} \\
& y^{\prime \prime}=\frac{-a \cos (\log x)}{x^2}+\frac{a \sin (\log x)}{x^2} \\
& -\frac{b \sin (\log x)}{x^2}-\frac{b \cos (\log x)}{x^2} \\
& x^2 y^{\prime \prime}=-[a \cos (\log x)+b \sin (\log x)] \\
& -[a \sin (\log x)+b \cos (\log x)] \\
& \Rightarrow \quad x^2 y^{\prime \prime}=-y-x y^{\prime} \\
& \Rightarrow \quad x^2 y^{\prime \prime}+x y^{\prime}=-y \\
&
\end{aligned}
$$
$$
\begin{aligned}
& y=a \cos (\log x)+b \sin (\log x) \\
& y^{\prime}=\frac{-a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x} \\
& y^{\prime \prime}=\frac{-a \cos (\log x)}{x^2}+\frac{a \sin (\log x)}{x^2} \\
& -\frac{b \sin (\log x)}{x^2}-\frac{b \cos (\log x)}{x^2} \\
& x^2 y^{\prime \prime}=-[a \cos (\log x)+b \sin (\log x)] \\
& -[a \sin (\log x)+b \cos (\log x)] \\
& \Rightarrow \quad x^2 y^{\prime \prime}=-y-x y^{\prime} \\
& \Rightarrow \quad x^2 y^{\prime \prime}+x y^{\prime}=-y \\
&
\end{aligned}
$$
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