MCQ
The solution of the differential equation ${\cos ^2}x\frac{{{d^2}y}}{{d{x^2}}} = 1$ is
  • A
    $y = \log \cos x + cx$
  • B
    $y = \log \sec x + {c_1}x + {c_2}$
  • C
    $y = \log \sec x - {c_1}x + {c_2}$
  • Both $(b)$ and $(c)$

Answer

Correct option: D.
Both $(b)$ and $(c)$
d
(d) ${\cos ^2}x\frac{{{d^2}y}}{{d{x^2}}} = 1$ ==> $\frac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}x$

On integrating, we get $\frac{{dy}}{{dx}} = \tan x \pm {c_1}$

Again integrating, we get $y = \log \sec x \pm {c_1}x \pm {c_2}$.

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