If $y=A \sin x+B \cos x$,then prove that $\frac{d^{2} y}{d x^{2}}+y=0$.

Given $y = A \sin x + B \cos x$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = A \cos x - B \sin x$.
Differentiating again with respect to $x$,we get:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(A \cos x - B \sin x) = -A \sin x - B \cos x$.
We can factor out $-1$ from the expression:
$\frac{d^{2}y}{dx^{2}} = -(A \sin x + B \cos x)$.
Since $y = A \sin x + B \cos x$,we substitute $y$ into the equation:
$\frac{d^{2}y}{dx^{2}} = -y$.
Rearranging the terms,we get:
$\frac{d^{2}y}{dx^{2}} + y = 0$.
Hence,the statement is proved.