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If y=x^(n-1)lnx then x^2y2+(3-2n)x y1 i...

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

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Given, `y=x^(n-1)lnx....(1)` `dy/dx=x^(n-2)+(n-1)x^(n-2)lnx`
`(d^2y)/(dx^2)=(n-2)x^(n-3)+(n-1)x^(n-2).1/x+(n-1)(n-2)x^(n-3)lnx`
`(d^2y)/(dx^2)=(n-2)x^(n-3)+(n-1)x^(n-3)+(n-1)(n-2)x^(n-3)lnx`
`x^2(d^2y)/(dx^2)=(n-2)x^(n-1)+(n-1)x^(n-1).+(n-1)(n-2)x^(n-1)lnx....(2)`
`(3-2n)dy/dx=(3-2n)x^(n-1)+(3-2n)(n-1)x^(n-2)lnx....(3)`
From (1),(2) and (3)
`x^2y_2+(3-2n)x y_1`=`0`
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