1. **State the problem:** Find the differential equation for the function $$y = A \sin 3x + B \cos 3x$$ where $A$ and $B$ are constants. 2. **Differentiate the function:** Compute the first derivative of $y$ with respect to $x$: $$y' = \frac{dy}{dx} = 3A \cos 3x - 3B \sin 3x$$ 3. **Differentiate again:** Compute the second derivative: $$y'' = \frac{d^2y}{dx^2} = -9A \sin 3x - 9B \cos 3x$$ 4. **Express $y''$ in terms of $y$:** Notice that $$y = A \sin 3x + B \cos 3x$$ so $$y'' = -9 (A \sin 3x + B \cos 3x) = -9y$$ 5. **Write the differential equation:** Rearranging, $$y'' + 9y = 0$$ **Final answer:** The differential equation satisfied by $$y = A \sin 3x + B \cos 3x$$ is $$y'' + 9y = 0$$