If `y=x^(n-1)logx` then `x^2y_2+(3-2n)x y_1` is equal to `-(n-1)^2y` (b) `(n-1)^2y` `-n^2y` (d) `n^2y`

Explanation

`y=x^{n-1} log x`
`
Rightarrow y_{1}=(n-1) x^{n-2} log x+frac{x^{n-1}}{x}`
`
Rightarrow y_{1}=frac{(n-1) x^{n-1} log x+x^{n-1}}{x}`
`
Rightarrow x y_{1}=(n-1) y+x^{n-1}`
`
Rightarrow x y_{2}+y_{1}=(n-1) y_{1}+(n-1) x^{n-2}`
`
Rightarrow x y_{2}+y_{1}=(n-1) y_{1}+frac{(n-1) x^{n-1}}{x}`
`
Rightarrow x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1) x^{n-1}`
`
Rightarrow x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1){x y_{1}-(n-1) y_{3}.`
`
...