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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Aicharya · Accepted Answer

We have, $\begin{aligned} & y=a \cos (\log x)+b \sin (\log x) \ & y^{\prime}=\frac{-a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x} \ & y^{\prime \prime}=\frac{-a \cos (\log x)}{x^2}+\frac{a \sin (\log x)}{x^2} \ & -\frac{b \sin (\log x)}{x^2}-\frac{b \cos (\log x)}{x^2} \ & x^2 y^{\prime \prime}=-[a \cos (\log x)+b \sin (\log x)] \ & -[a \sin (\log x)+b \cos (\log x)] \ & \Rightarrow \quad x^2 y^{\prime \prime}=-y-x y^{\prime} \ & \Rightarrow \quad x^2 y^{\prime \prime}+x y^{\prime}=-y \ & \end{aligned}$

If $y=a \cos (\log x)+b \sin (\log x)$, where $a, b$ are parameters, then $x^2 y^{\prime \prime}+x…

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