If

It is given that, $y=5 \cos x-3 \sin x$

Then

$\frac{d y}{d x}=\frac{d}{d x}(5 \cos x)-\frac{d}{d x}(3 \sin x)=5 \frac{d}{d x}(\cos x)-3 \frac{d}{d x}(\sin x)$

$=5(-\sin x)-3 \cos x=-(5 \sin x+3 \cos x)$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[-(5 \sin x+3 \cos x)]$

$=-\left[5 \cdot \frac{d}{d x}(\sin x)+3 \cdot \frac{d}{d x}(\cos x)\right]$

$=-[5 \cos x+3(-\sin x)]$

$=-[5 \cos x-3 \sin x]$

$=-y$

$\therefore \frac{d^{2} y}{d x^{2}}+y=0$

Hence, proved.