MCQ
If $x = \log p$ and $y = {1 \over p}$, then
- A${{{d^2}y} \over {d{x^2}}} - 2p = 0$
- B${{{d^2}y} \over {d{x^2}}} + y = 0$
- ✓${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} = 0$
- D${{{d^2}y} \over {d{x^2}}} - {{dy} \over {dx}} = 0$
$ \Rightarrow \frac{{dy}}{{dx}} = - {e^{ - x}}$ and $\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - x}};\,\,\,$
$\therefore \frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} = 0$.
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