$y=A \cos \alpha x+B \sin \alpha x$
Let us differentiate this w.r.t x on both sides
$\Rightarrow \large \frac{dy}{dx}$$=-\alpha A \sin \alpha x+\alpha B \cos \alpha x$
Again differentiating w.r.t x we get
$\Rightarrow \large \frac{d^2y}{dx^2}$$=-\alpha ^2 A \cos \alpha x-\alpha ^2B \sin \alpha x$
$\large \frac{d^2y}{dx^2}$$=-\alpha ^2(A \cos \alpha x+B\sin \alpha x)$
But $ A \cos \alpha x+ B \sin \alpha x=y$
$\therefore \large \frac{d^2y}{dx^2}$$=-\alpha ^2y$
$\Rightarrow \large \frac{d^2y}{dx^2}$$+\alpha ^2y=0$
Hence the correct option is B