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

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

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

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

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