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