If y = frac{A}{x} + B x^2, then x^2 frac{d^2 | vedclass

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(A) Given the equation: $y = A x^{-1} + B x^2$. First, differentiate $y$ with respect to $x$: $\frac{d y}{d x} = -A x^{-2} + 2 B x$. Next, differentia

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$2 y$

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(A) Given the equation: $y = A x^{-1} + B x^2$. First, differentiate $y$ with respect to $x$: $\frac{d y}{d x} = -A x^{-2} + 2 B x$. Next, differentiate again with respect to $x$ to find the second derivative: $\frac{d^2 y}{d x^2} = 2 A x^{-3} + 2 B$. Now, multiply by $x^2$: $x^2 \frac{d^2 y}{d x^2} = x^2 (2 A x^{-3} + 2 B) = 2 A x^{-1} + 2 B x^2$. Factor out $2$: $x^2 \frac{d^2 y}{d x^2} = 2 (A x^{-1} + B x^2)$. Since $y = A x^{-1} + B x^2$, we substitute $y$ back into the expression: $x^2 \frac{d^2 y}{d x^2} = 2 y$.

If $y = \frac{A}{x} + B x^2$, then $x^2 \frac{d^2 y}{d x^2} =$

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