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

Answers (1)

Answer:   Option (c)

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)

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