Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 24

(a)

Hint:

We must know about the derivative of logarithm.

Given:

$y=x^{n-1} \log x$

Explanation:

$y=x^{n-1} \log x$

$\\ \frac{d y}{d x}=y_{1}=(n-1) x^{n-2} \log x+\frac{x^{n-1}}{x} \\$

\begin{aligned} & &y_{1}=\frac{(n-1) x^{n-1} \log x+x^{n-1}}{x} \end{aligned}

$x y_{1}=(n-1) y+x^{n-1} \\$                                                                                                     $\left[\because y=x^{n-1} \log x\right]$

$x y_{2}+y_{1}=(n-1) y_{1}+(n-1) x^{n-2}$

\begin{aligned} & \\ &x y_{2}+y_{1}=(n-1) y_{1}+\frac{(n-1) x^{n-1}}{x} \end{aligned}

$x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1) x^{n-1}$

$\\ x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1)\left[x y_{1}-(n-1) y\right] \\$

$x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1) x y_{1}-(n-1)^{2} y \\$

\begin{aligned} & &x^{2} y_{2}+x y_{1}=2 x(n-1) y_{1}-(n-1)^{2} y \end{aligned}

$x^{2} y_{2}+x y_{1}-2 x(n-1) y_{1}=-(n-1)^{2} y \\$

$x^{2} y_{2}+x y_{1}(1-2 n+2)=-(n-1)^{2} y \\$

\begin{aligned} & &x^{2} y_{2}+(3-2 n) x y_{1}=-(n-1)^{2} y \end{aligned}