y=acosx+bsinx where a, b are arbitrary constants is a solution of the…

The correct answer is Option (3) → $\frac{d^2y}{dx^2}+y=0$

$y=a\cos x+b\sin x$

$\frac{dy}{dx}=-a\sin x+b\cos x$

$\frac{d^2y}{dx^2}=-a\cos x-b\sin x$

so $\frac{d^2y}{dx^2}=-y⇒\frac{d^2y}{dx^2}+y=0$