1. **Problem statement:** Given the function $y = e^{-x}(\cos 2x + \sin 2x)$, find the values of $p$ and $q$ such that the differential equation $$\frac{d^2y}{dx^2} + p \frac{dy}{dx} + qy = 0$$ holds. 2. **Step 1: Differentiate $y$ once to find $\frac{dy}{dx}$.** Use the product rule: $$\frac{dy}{dx} = \frac{d}{dx}\left(e^{-x}\right)(\cos 2x + \sin 2x) + e^{-x} \frac{d}{dx}(\cos 2x + \sin 2x)$$ Calculate derivatives: $$\frac{d}{dx} e^{-x} = -e^{-x}$$ $$\frac{d}{dx} \cos 2x = -2 \sin 2x$$ $$\frac{d}{dx} \sin 2x = 2 \cos 2x$$ So, $$\frac{dy}{dx} = -e^{-x}(\cos 2x + \sin 2x) + e^{-x}(-2 \sin 2x + 2 \cos 2x)$$ Simplify inside the parentheses: $$- (\cos 2x + \sin 2x) + (-2 \sin 2x + 2 \cos 2x) = (-\cos 2x - \sin 2x) + (-2 \sin 2x + 2 \cos 2x) = (-\cos 2x + 2 \cos 2x) + (-\sin 2x - 2 \sin 2x) = \cos 2x - 3 \sin 2x$$ Therefore, $$\frac{dy}{dx} = e^{-x}(\cos 2x - 3 \sin 2x)$$ 3. **Step 2: Differentiate $\frac{dy}{dx}$ to find $\frac{d^2y}{dx^2}$.** Again use product rule: $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(e^{-x}\right)(\cos 2x - 3 \sin 2x) + e^{-x} \frac{d}{dx}(\cos 2x - 3 \sin 2x)$$ Calculate derivatives: $$\frac{d}{dx} e^{-x} = -e^{-x}$$ $$\frac{d}{dx} \cos 2x = -2 \sin 2x$$ $$\frac{d}{dx} (-3 \sin 2x) = -3 \times 2 \cos 2x = -6 \cos 2x$$ So, $$\frac{d^2y}{dx^2} = -e^{-x}(\cos 2x - 3 \sin 2x) + e^{-x}(-2 \sin 2x - 6 \cos 2x)$$ Simplify inside the parentheses: $$-(\cos 2x - 3 \sin 2x) + (-2 \sin 2x - 6 \cos 2x) = (-\cos 2x + 3 \sin 2x) + (-2 \sin 2x - 6 \cos 2x) = (-\cos 2x - 6 \cos 2x) + (3 \sin 2x - 2 \sin 2x) = -7 \cos 2x + \sin 2x$$ Therefore, $$\frac{d^2y}{dx^2} = e^{-x}(-7 \cos 2x + \sin 2x)$$ 4. **Step 3: Substitute $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the differential equation:** $$\frac{d^2y}{dx^2} + p \frac{dy}{dx} + q y = 0$$ Substitute: $$e^{-x}(-7 \cos 2x + \sin 2x) + p e^{-x}(\cos 2x - 3 \sin 2x) + q e^{-x}(\cos 2x + \sin 2x) = 0$$ Divide both sides by $e^{-x}$ (never zero): $$-7 \cos 2x + \sin 2x + p(\cos 2x - 3 \sin 2x) + q(\cos 2x + \sin 2x) = 0$$ Group terms: $$( -7 + p + q ) \cos 2x + (1 - 3p + q) \sin 2x = 0$$ 5. **Step 4: Set coefficients of $\cos 2x$ and $\sin 2x$ to zero for the equation to hold for all $x$:** $$-7 + p + q = 0$$ $$1 - 3p + q = 0$$ 6. **Step 5: Solve the system of equations:** From first: $$q = 7 - p$$ Substitute into second: $$1 - 3p + (7 - p) = 0$$ $$1 - 3p + 7 - p = 0$$ $$8 - 4p = 0$$ $$4p = 8$$ $$p = 2$$ Then, $$q = 7 - 2 = 5$$ **Final answer:** $$p = 2, \quad q = 5$$