A) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2p=0\]
B) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=0\]
C) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}=0\]
D) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-\frac{dy}{dx}=0\]
Correct Answer: C
Solution :
\[x=\log p\Rightarrow p={{e}^{x}}\Rightarrow y={{e}^{-x}}\] \[\Rightarrow \frac{dy}{dx}=-{{e}^{-x}}\] 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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