1. **Stating the problem:** Solve the differential equation given by $$(9D^2 - 6D + 1)y = 0,$$ where $D$ represents the differential operator $\frac{d}{dx}$. 2. **Understanding the operator:** The operator $D$ acts as $D y = \frac{dy}{dx}$ and $D^2 y = \frac{d^2 y}{dx^2}$. The equation can be rewritten as a linear differential equation: $$9 \frac{d^2 y}{dx^2} - 6 \frac{dy}{dx} + y = 0.$$ 3. **Characteristic equation:** Replace $D$ by $r$ to find the characteristic polynomial: $$9r^2 - 6r + 1 = 0.$$ 4. **Solving the quadratic:** Use the quadratic formula: $$r = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 9 \times 1}}{2 \times 9} = \frac{6 \pm \sqrt{36 - 36}}{18} = \frac{6 \pm 0}{18} = \frac{6}{18} = \frac{1}{3}.$$ 5. **Roots:** The characteristic equation has a repeated root $r = \frac{1}{3}$. 6. **General solution:** For repeated roots $r$, the general solution is: $$y = (C_1 + C_2 x) e^{rx} = (C_1 + C_2 x) e^{\frac{x}{3}}.$$ **Final answer:** $$y = (C_1 + C_2 x) e^{\frac{x}{3}}.$$