1. **State the problem:** We are given the differential equation $$\frac{dy}{dx} = \frac{1}{2y + 1}$$ with the initial condition $$y(0) = 0$$. We need to find the particular solution $$y = f(x)$$ and determine the domain where this solution is valid. 2. **Separate variables:** Rewrite the equation to separate variables: $$\frac{dy}{dx} = \frac{1}{2y + 1} \implies (2y + 1) dy = dx$$ 3. **Integrate both sides:** $$\int (2y + 1) dy = \int dx$$ 4. **Perform the integration:** $$\int (2y + 1) dy = \int 2y dy + \int 1 dy = y^2 + y + C_1$$ $$\int dx = x + C_2$$ 5. **Combine constants:** Let $$C = C_2 - C_1$$, then $$y^2 + y = x + C$$ 6. **Apply initial condition:** When $$x=0$$, $$y=0$$, so $$0^2 + 0 = 0 + C \implies C = 0$$ 7. **Rewrite implicit solution:** $$y^2 + y = x$$ 8. **Solve quadratic for y:** $$y^2 + y - x = 0$$ Use quadratic formula: $$y = \frac{-1 \pm \sqrt{1 + 4x}}{2}$$ 9. **Choose correct branch:** Since $$y(0) = 0$$, substitute $$x=0$$: $$y = \frac{-1 \pm \sqrt{1}}{2} = \frac{-1 \pm 1}{2}$$ Possible values: $$0$$ or $$-1$$. We want $$y(0) = 0$$, so choose the positive root: $$y = \frac{-1 + \sqrt{1 + 4x}}{2}$$ 10. **Determine domain:** The expression under the square root must be non-negative: $$1 + 4x \geq 0 \implies x \geq -\frac{1}{4}$$ 11. **Check domain for validity:** The solution is valid for $$x > -\frac{1}{4}$$ because at $$x = -\frac{1}{4}$$ the derivative becomes undefined (denominator zero). **Final answer:** $$f(x) = \frac{-1 + \sqrt{1 + 4x}}{2}$$ valid for $$x > -\frac{1}{4}$$. This corresponds to option D.