1. The problem asks us to verify if the function \(y = e^{-2x} (\cos 3x + \sin 3x)\) is a solution to the differential equation \(\frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + 13y = 0\). 2. First, find the first derivative \(\frac{dy}{dx}\): \[y = e^{-2x} (\cos 3x + \sin 3x)\] Using the product rule: \[\frac{dy}{dx} = e^{-2x} \frac{d}{dx}(\cos 3x + \sin 3x) + (\cos 3x + \sin 3x) \frac{d}{dx}(e^{-2x})\] Calculate each derivative: \[\frac{d}{dx}(\cos 3x) = -3 \sin 3x, \quad \frac{d}{dx}(\sin 3x) = 3 \cos 3x, \quad \frac{d}{dx}(e^{-2x}) = -2 e^{-2x}\] So: \[\frac{dy}{dx} = e^{-2x} (-3 \sin 3x + 3 \cos 3x) + (\cos 3x + \sin 3x)(-2 e^{-2x})\] Simplify: \[\frac{dy}{dx} = e^{-2x} (-3 \sin 3x + 3 \cos 3x - 2 \cos 3x - 2 \sin 3x) = e^{-2x} ((3 - 2) \cos 3x + (-3 - 2) \sin 3x) = e^{-2x} (\cos 3x - 5 \sin 3x)\] 3. Next, find the second derivative \(\frac{d^2 y}{dx^2}\): Again use the product rule: \[\frac{d^2 y}{dx^2} = \frac{d}{dx}[e^{-2x} (\cos 3x - 5 \sin 3x)] = e^{-2x} \frac{d}{dx}(\cos 3x - 5 \sin 3x) + (\cos 3x - 5 \sin 3x) \frac{d}{dx}(e^{-2x})\] Calculate the derivatives inside: \[\frac{d}{dx}(\cos 3x) = -3 \sin 3x, \quad \frac{d}{dx}(-5 \sin 3x) = -15 \cos 3x, \quad \frac{d}{dx}(e^{-2x}) = -2 e^{-2x}\] So: \[\frac{d^2 y}{dx^2} = e^{-2x} (-3 \sin 3x - 15 \cos 3x) + (\cos 3x - 5 \sin 3x)(-2 e^{-2x})\] Simplify terms: \[\frac{d^2 y}{dx^2} = e^{-2x} (-3 \sin 3x - 15 \cos 3x - 2 \cos 3x + 10 \sin 3x) = e^{-2x}((-15 - 2) \cos 3x + (-3 + 10) \sin 3x) = e^{-2x} (-17 \cos 3x + 7 \sin 3x)\] 4. Substitute \(y\), \(\frac{dy}{dx}\), and \(\frac{d^2 y}{dx^2}\) into the differential equation: \[\frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + 13 y = e^{-2x}(-17 \cos 3x + 7 \sin 3x) + 4 e^{-2x}(\cos 3x - 5 \sin 3x) + 13 e^{-2x} (\cos 3x + \sin 3x)\] Combine the terms inside the brackets: \[(-17 \cos 3x + 7 \sin 3x) + 4(\cos 3x - 5 \sin 3x) + 13 (\cos 3x + \sin 3x) = (-17 + 4 + 13) \cos 3x + (7 - 20 + 13) \sin 3x = 0 \cos 3x + 0 \sin 3x = 0 \] 5. Since the expression equals zero, the function satisfies the differential equation. **Final answer:** Yes, \(y = e^{-2x} (\cos 3x + \sin 3x)\) is a solution to \(\frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + 13 y = 0\).