1. **Problem Statement:** Solve the differential equation $$\left(D^6 - 64\right)y = 0$$ where $D$ denotes differentiation with respect to $x$. 2. **Characteristic Equation:** Replace $D$ by $r$ to get the characteristic polynomial: $$r^6 - 64 = 0$$ 3. **Solve the Characteristic Equation:** Rewrite as: $$r^6 = 64 = 2^6$$ The sixth roots of $64$ are: $$r = 2 e^{i \frac{2\pi k}{6}} = 2 e^{i \frac{\pi k}{3}}, \quad k=0,1,2,3,4,5$$ 4. **Find the Roots:** For $k=0$: $r=2$ For $k=1$: $r=2 e^{i \pi/3} = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 1 + i \sqrt{3}$ For $k=2$: $r=2 e^{i 2\pi/3} = 2(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = -1 + i \sqrt{3}$ For $k=3$: $r=2 e^{i \pi} = 2(-1) = -2$ For $k=4$: $r=2 e^{i 4\pi/3} = 2(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -1 - i \sqrt{3}$ For $k=5$: $r=2 e^{i 5\pi/3} = 2(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 1 - i \sqrt{3}$ 5. **Group Real and Complex Roots:** Real roots: $r=2$ and $r=-2$ Complex conjugate pairs: $$1 \pm i \sqrt{3}$$ and $$-1 \pm i \sqrt{3}$$ 6. **General Solution Form:** For real roots $r=a$, solutions are $e^{ax}$. For complex roots $\alpha \pm i \beta$, solutions are: $$e^{\alpha x} (C_1 \cos \beta x + C_2 \sin \beta x)$$ 7. **Apply to Our Roots:** - For $r=2$: solution $C_5 e^{2x}$ - For $r=-2$: solution $C_6 e^{-2x}$ - For $r=1 \pm i \sqrt{3}$: $\alpha=1$, $\beta=\sqrt{3}$, solution: $$e^{x} (C_3 \cos(\sqrt{3} x) + C_4 \sin(\sqrt{3} x))$$ - For $r=-1 \pm i \sqrt{3}$: $\alpha=-1$, $\beta=\sqrt{3}$, solution: $$e^{-x} (C_1 \cos(\sqrt{3} x) + C_2 \sin(\sqrt{3} x))$$ 8. **Final General Solution:** $$y = e^{-x} (C_1 \cos(\sqrt{3} x) + C_2 \sin(\sqrt{3} x)) + e^{x} (C_3 \cos(\sqrt{3} x) + C_4 \sin(\sqrt{3} x)) + C_5 e^{2x} + C_6 e^{-2x}$$ **Answer:** The first option matches this solution.