MCQ
The solution of $\frac{{{d^2}y}}{{d{x^2}}} = \cos x - \sin x$is
  • $y = - \cos x + \sin x + {c_1}x + {c_2}$
  • B
    $y = - \cos x - \sin x + {c_1}x + {c_2}$
  • C
    $y = \cos x - \sin x + {c_1}{x^2} + {c_2}x$
  • D
    $y = \cos x + \sin x + {c_1}{x^2} + {c_2}x$

Answer

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

$\frac{{dy}}{{dx}} = \sin x + \cos x + {c_1}$

Again $y = - \cos x + \sin x + {c_1}x + {c_2}$.

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