Explanation

(a)
Here, $y = A\cos{\omega t + B\sin{\omega t}}$ ….(i)
On differentiating $w.r.t.t,$ we get
$\frac{dy}{dt} = - \omega A\sin{\omega t + \omega B\cos{\omega t}}$
Again, on differentiating $w.r.t.t,$ we get
$\frac{d^{2}y}{dt^{2}} = - \omega^{2}\ A\cos{\omega t - \omega^{2}\ B\sin{\omega t}}$
$\Rightarrow \frac{d^{2}y}{dt^{2}} = - \omega^{2}(A\cos{\omega t - \ B\sin{\omega t)}}$
$\therefore y_{2} = {- \omega}^{2}y$ [from Eq. (i)]