Write the correct alternative in the following:

Question:

Write the correct alternative in the following:

If $y=a \cos \left(\log _{e} x\right)+b \sin \left(\log _{e} x\right)$, then $x^{2} y_{2}+x y_{1}=$

A. 0

B. $y$

C. $-y$

D. none of these

Solution:

Given:

$y=a \cos \left(\log _{e} x\right)+b \sin \left(\log _{e} x\right)$

$\frac{d y}{d x}=-a \sin \left(\log _{e} x\right) \frac{1}{x}+b \cos \left(\log _{e} x\right) \frac{1}{x}$

$x y_{1}=-a \sin \left(\log _{e} x\right)+b \cos \left(\log _{e} x\right)$

$\frac{d^{2} y}{d x^{2}}=-a \cos \left(\log _{e} x\right) \frac{1}{x^{2}}+\frac{1}{x^{2}} a \sin \left(\log _{e} x\right)-b \sin \left(\log _{e} x\right) \frac{1}{x^{2}}+b \cos \left(\log _{e} x\right) \frac{1}{x^{2}}$

$x^{2} y_{2}=-a \cos \left(\log _{e} x\right)+a \sin \left(\log _{e} x\right)-b \sin \left(\log _{e} x\right)-b \cos \left(\log _{e} x\right)$

$=-a \sin \left(\log _{e} x\right)-b \cos \left(\log _{e} x\right)$

$=-y$

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