Write the correct alternative in the following:
Question:

Write the correct alternative in the following:

If $y=x^{n-1} \log x$, then $x^{2} y_{2}+(3-2 n) x y_{1}$ is equal to

A. $-(n-1)^{2} y$

B. $(n-1)^{2} y$

C. $-n^{2} y$

D. $n^{2} y$

Solution:

Given:

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

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

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

$=x^{n-2}[(n-1) \log x+1]$

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

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

$(3-2 n) x y_{1}=(3-2 n)\left[(n-1) y+x^{n-1}\right]$

$=\left(3 n-3-2 n^{2}+2 n\right) y+3 x^{n-1}-2 n x^{n-1}$ (1)

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=(\mathrm{n}-1)(\mathrm{n}-2) \mathrm{x}^{\mathrm{n}-3} \log \mathrm{x}+\frac{1}{\mathrm{x}}(\mathrm{n}-1) \mathrm{x}^{\mathrm{n}-2}+(\mathrm{n}-2) \mathrm{x}^{\mathrm{n}-3}$

$=(n-1)(n-2) x^{n-3} \log x+(n-1) x^{n-3}+(n-2) x^{n-3}$

$=x^{n-3}[(n-1)(n-2) \log x+(n-1)+(n-2)]$

$x^{2} y_{2}=x^{n-1}[(n-1)(n-2) \log x+(2 n-3)]$

$=\left(n^{2}-3 n+2\right) y+2 n x^{n-1}-3 x^{n-1}(2)$

$x^{2} y_{2}+(3-2 n) x y_{1}$

$=\left(n^{2}-3 n+2\right) y+2 n x^{n-1}-3 x^{n-1}+\left(3 n-3-2 n^{2}+2 n\right) y+3 x^{n-1}-2 n x^{n-1}$

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

$=-(n-1)^{2} y$

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