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

We have, $y = 5\cos x - 3\sin x$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = 5( - \sin x) - 3(\cos x) = - 5\sin x - 3\cos x$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} = - 5\cos x - 3( - \sin x) = - 5\cos x + 3\sin x$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} + y = - 5\cos x + 3\sin x + 5\cos x - 3\sin x = 0$
Hence proved.