MCQ
If $y = A\cos nx + B\sin nx,$ then ${{{d^2}y} \over {d{x^2}}} = $
  • A
    ${n^2}y$
  • B
    $ - y$
  • $ - {n^2}y$
  • D
    None of these

Answer

Correct option: C.
$ - {n^2}y$
c
(c) $y = A\cos (nx) + B\sin (nx)$

$\therefore dy/dx = - nA\sin (nx) + nB\cos (nx)$

Again $\frac{{{d^2}y}}{{d{x^2}}} = - {n^2}A\cos (nx) - {n^2}B\sin (nx)$

$ = - {n^2}[A\cos (nx) + B\sin (nx)]$

==> $\frac{{{d^2}y}}{{d{x^2}}} = - {n^2}y$.

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