If $x = \log p$ and $y = \frac{1}{p}$, then
A$\frac{d^2y}{dx^2} - 2p = 0$
B$\frac{d^2y}{dx^2} + y = 0$
C$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$
D$\frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$
Answer & Solution
Correct answer: C. $\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$
Given $x = \log p$, we have $$\frac{dx}{dp} = \frac{1}{p}.$$ Also, since $y = \frac{1}{p}$, $$\frac{dy}{dp} = -\frac{1}{p^2}.$$ Therefore, $$\frac{dy}{dx} = \frac{\frac{dy}{dp}}{\frac{dx}{dp}} = -\frac{1}{p} = -y.$$ Differentiate again with respect to $x$: $$\frac{d^2y}{dx^2} = -\frac{dy}{dx}.$$ Hence, $$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0.$$ Comparing with the given options, this matches option $\mathrm{(C)}$.
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