A) \[y=\log x+{{c}_{1}}x+{{c}_{2}}\]
B) \[y=-\log x+{{c}_{1}}x+{{c}_{2}}\]
C) \[y=-\frac{1}{x}+{{c}_{1}}x+{{c}_{2}}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\frac{1}{{{x}^{2}}}\]. Now integrating both sides, we get \[\frac{dy}{dx}=\frac{1}{x}+{{c}_{1}}\] Þ \[y=\log x+{{c}_{1}}x+{{c}_{2}}\].You need to login to perform this action.
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