If x = p and y = 1 p , then - Vidyadip

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
${{{d^2}y} \over {d{x^2}}} - {{dy} \over {dx}} = 0$

✓

Answer

Correct option: C.

${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} = 0$

c
(c) $x = \log p \Rightarrow p = {e^x} \Rightarrow y = {e^{ - x}}$

$ \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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