Verify that the given function $y=x \sin 3x$ is a solution of the differential equation $\frac{d^{2}y}{dx^{2}}+9y-6 \cos 3x=0$.

(A) Given function: $y = x \sin 3x$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \sin 3x + x \cdot \frac{d}{dx}(\sin 3x)$
$\frac{dy}{dx} = 1 \cdot \sin 3x + x \cdot (\cos 3x \cdot 3) = \sin 3x + 3x \cos 3x$
Differentiating again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\sin 3x) + 3 \cdot \frac{d}{dx}(x \cos 3x)$
$\frac{d^{2}y}{dx^{2}} = 3 \cos 3x + 3 [1 \cdot \cos 3x + x \cdot (-\sin 3x \cdot 3)]$
$\frac{d^{2}y}{dx^{2}} = 3 \cos 3x + 3 \cos 3x - 9x \sin 3x = 6 \cos 3x - 9x \sin 3x$
Now,substitute $\frac{d^{2}y}{dx^{2}}$ and $y$ into the $L.H.S.$ of the differential equation:
$L.H.S. = \frac{d^{2}y}{dx^{2}} + 9y - 6 \cos 3x$
$= (6 \cos 3x - 9x \sin 3x) + 9(x \sin 3x) - 6 \cos 3x$
$= 6 \cos 3x - 6 \cos 3x - 9x \sin 3x + 9x \sin 3x$
$= 0 = R.H.S.$
Since $L.H.S. = R.H.S.$,the given function is a solution of the differential equation.