Problem 05 | Elimination of Arbitrary Constants

$x = A \sin (\omega t + B)$   ←   equation (1)

$\dfrac{dx}{dt} = \omega A \, \cos (\omega t + B)$

$\dfrac{d^2x}{dt^2} = -\omega^2 A \, \sin (\omega t + B)$
 

From equation (1), A sin (ωt + B) = x
$\dfrac{d^2x}{dt^2} = -\omega^2 x$

$\dfrac{d^2x}{dt^2} + \omega^2 x = 0$           answer