The differential equation for y = A cos αx + B sin αx where A and B are arbitary constants is

The correct option is d²y/dx² + α²y = 0.

Let us find the differential equation by eliminating the arbitrary constants $A$ and $B$ from the given equation:
$y=A\cos \left(\alpha x\right)+B\sin \left(\alpha x\right)$          --- (Equation 1)

Differentiating Equation 1 with respect to $x$, we get:
$\frac{dy}{dx}=-A\alpha \sin \left(\alpha x\right)+B\alpha \cos \left(\alpha x\right)$

Differentiating once again with respect to $x$, we obtain:
$\frac{d^{2}y}{d{x}^2}=-A{\alpha }^2\cos \left(\alpha x\right)-B{\alpha }^2\sin \left(\alpha x\right)$

Factoring out $-{\alpha }^2$ from the right-hand side, we get:
$\frac{d^{2}y}{d{x}^2}=-{\alpha }^2\left[A\cos \left(\alpha x\right)+B\sin \left(\alpha x\right)\right]$

Substituting the value of $y$ from Equation 1 into the brackets:
$\frac{d^{2}y}{d{x}^2}=-{\alpha }^{2}y$

Rearranging the equation to standard form gives:
$\frac{d^{2}y}{d{x}^2}+{\alpha }^{2}y=0$