Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question Question 12

Hint:

We must know about the derivative of  $\cos x$ and logarithm.

Given:

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

Explanation:

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

Differentiate both side with respect to  $x$

$\frac{d y}{d x}=\frac{a[-\sin (\log x)]}{x}-\frac{b[\cos (\log x)]}{x} \\$

$\frac{d y}{d x}=\frac{-[a \sin (\log x)+b \cos (\log x)]}{x} \$

\begin{aligned} \ &x \frac{d y}{d x}=-[a \sin (\log x)+b \cos (\log x)] \end{aligned}

Again differentiating with respect to  $x$ ,

\begin{aligned} &x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\left[\frac{a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x}\right] \\ \end{aligned}

\begin{aligned} &x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\left[\frac{a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x}\right] \\ \end{aligned}

$x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=\frac{-y}{x} \\$

$x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{y}{x}=0$

\begin{aligned} \\ &x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0 \end{aligned}    or

$x^{2}y_{2}+xy_{1}+y=0$

Hence option the value of   $x^{2}y_{2}+xy_{1}$   is $(-y)$