MCQ
The solution of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{{x^2}}}$ is
  • $y = \log x + {c_1}x + {c_2}$
  • B
    $y = - \log x + {c_1}x + {c_2}$
  • C
    $y = - \frac{1}{x} + {c_1}x + {c_2}$
  • D
    None of these

Answer

Correct option: A.
$y = \log x + {c_1}x + {c_2}$
a
(a) $\frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{{x^2}}}$. Now integrating both sides, we get

$\frac{{dy}}{{dx}} = \frac{1}{x} + {c_1}$ ==> $y = \log x + {c_1}x + {c_2}$.

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