If y = a{x^{n + 1}} + b{x^{ - n}},then {x^2}f | vedclass

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(B) Given $y = a{x^{n + 1}} + b{x^{ - n}}$.Differentiating with respect to $x$:$\frac{dy}{dx} = a(n + 1){x^n} + b(-n){x^{ - n - 1}} = a(n + 1){x^n} -

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$n(n + 1)y$

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(B) Given $y = a{x^{n + 1}} + b{x^{ - n}}$.Differentiating with respect to $x$:$\frac{dy}{dx} = a(n + 1){x^n} + b(-n){x^{ - n - 1}} = a(n + 1){x^n} - nb{x^{ - n - 1}}$.Differentiating again with respect to $x$:$\frac{{{d^2}y}}{{d{x^2}}} = a(n + 1)n{x^{n - 1}} - nb(-n - 1){x^{ - n - 2}}$$\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)a{x^{n - 1}} + n(n + 1)b{x^{ - n - 2}}$.Now,multiply by ${x^2}$:${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)a{x^{n + 1}} + n(n + 1)b{x^{ - n}}$${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)(a{x^{n + 1}} + b{x^{ - n}})$.Since $y = a{x^{n + 1}} + b{x^{ - n}}$,we get:${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)y$.

If $y = a{x^{n + 1}} + b{x^{ - n}}$,then ${x^2}\frac{{{d^2}y}}{{d{x^2}}} = $

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