If $y=3 \cos (\log x)+4 \sin (\log x)$, then $x^2 y_2+x y_1=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{y}=3 \cos (\log \mathrm{x})+4 \sin (\log \mathrm{x}) \ & \Rightarrow \mathrm{y}_1=-3 \sin (\log \mathrm{x}) \cdot \frac{1}{\mathrm{x}}+4 \cos (\log \mathrm{x}) \cdot \frac{1}{\mathrm{x}} \ & \Rightarrow \mathrm{xy}_1=-3 \sin (\log \mathrm{x})+4 \cos (\log \mathrm{x}) \ & \Rightarrow 1 \cdot \mathrm{y}_1+\mathrm{x} \cdot \mathrm{y}_2=-3 \cos (\log \mathrm{x}) \cdot \frac{1}{\mathrm{x}}-4 \sin (\log \mathrm{x}) \cdot \frac{1}{\mathrm{x}} \ & \Rightarrow \mathrm{xy}_1+\mathrm{x}^2 \mathrm{y}_2=-3 \cos (\log \mathrm{x})-4 \sin (\log \mathrm{x})=-\mathrm{y}\end{aligned}$

If $y=3 \cos (\log x)+4 \sin (\log x)$, then $x^2 y_2+x y_1=$

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