The differential equation for which $y = a \cos x + b \sin x$ is a solution, is

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

The correct answer is Option (1) → $\frac{d^2y}{dx^2} + y = 0$ ##

Given that, $y = a \cos x + b \sin x$

To get differential equation we have to eliminate $a$ and $b$.

On differentiating both sides w.r.t. $x$, we get

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

Again, differentiating w.r.t. $x$, we get

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

$\Rightarrow \frac{d^2y}{dx^2} = -y \quad [∵ y = a \cos x + b \sin x]$

$\Rightarrow \frac{d^2y}{dx^2} + y = 0$