$x = A \sin (\omega t + B)$ ← 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