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

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(A) Given $y = a x^{n+1} + b x^{-n}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = a(n+1) x^n - bn x^{-n-1}$.Now,differentiate again with r

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$n(n+1) y$

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(A) Given $y = a x^{n+1} + b x^{-n}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = a(n+1) x^n - bn x^{-n-1}$.Now,differentiate again with respect to $x$ to find the second derivative:$\frac{d^2 y}{dx^2} = a(n+1)n x^{n-1} - bn(-n-1) x^{-n-2}$.$\frac{d^2 y}{dx^2} = a n(n+1) x^{n-1} + b n(n+1) x^{-n-2}$.Factor out $n(n+1)$:$\frac{d^2 y}{dx^2} = n(n+1) [a x^{n-1} + b x^{-n-2}]$.Multiply both sides by $x^2$:$x^2 \frac{d^2 y}{dx^2} = n(n+1) [a x^{n+1} + b x^{-n}]$.Since $y = a x^{n+1} + b x^{-n}$,we substitute $y$ into the equation:$x^2 \frac{d^2 y}{dx^2} = n(n+1) y$.

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

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