If $y=5 \cos x-3 \sin x,$ prove that $\frac{d^{2} y}{d x^{2}}+y=0.$

Given that,$y=5 \cos x-3 \sin x.$
Differentiating with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(5 \cos x) - \frac{d}{d x}(3 \sin x) = -5 \sin x - 3 \cos x.$
Differentiating again with respect to $x$:
$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x}(-5 \sin x - 3 \cos x) = -5 \cos x - 3(-\sin x) = -5 \cos x + 3 \sin x.$
Factor out $-1$:
$\frac{d^{2} y}{d x^{2}} = -(5 \cos x - 3 \sin x).$
Since $y = 5 \cos x - 3 \sin x,$ we have:
$\frac{d^{2} y}{d x^{2}} = -y.$
Therefore,$\frac{d^{2} y}{d x^{2}} + y = 0.$
Hence,proved.