1. The problem states that we have a differential equation given by $$y'(x) = \frac{x}{3} + \frac{y}{6}$$ and a slope field representing this equation. 2. This is a first-order linear differential equation of the form $$y' = ay + bx$$ where $$a = \frac{1}{6}$$ and $$b = \frac{1}{3}$$. 3. To solve it, we use the integrating factor method. The integrating factor $$\mu(x)$$ is given by: $$\mu(x) = e^{\int a \, dx} = e^{\int \frac{1}{6} \, dx} = e^{\frac{x}{6}}$$ 4. Multiply both sides of the differential equation by $$\mu(x)$$: $$e^{\frac{x}{6}} y' = e^{\frac{x}{6}} \left( \frac{x}{3} + \frac{y}{6} \right)$$ 5. The left side is the derivative of $$y e^{\frac{x}{6}}$$: $$\frac{d}{dx} \left( y e^{\frac{x}{6}} \right) = e^{\frac{x}{6}} \frac{x}{3}$$ 6. Integrate both sides with respect to $$x$$: $$y e^{\frac{x}{6}} = \int e^{\frac{x}{6}} \frac{x}{3} \, dx + C$$ 7. To integrate $$\int e^{\frac{x}{6}} \frac{x}{3} \, dx$$, use integration by parts: Let $$u = x$$, $$dv = \frac{1}{3} e^{\frac{x}{6}} dx$$ Then $$du = dx$$, $$v = 2 e^{\frac{x}{6}}$$ 8. Applying integration by parts: $$\int \frac{x}{3} e^{\frac{x}{6}} dx = uv - \int v du = 2x e^{\frac{x}{6}} - 2 \int e^{\frac{x}{6}} dx$$ 9. The remaining integral is: $$\int e^{\frac{x}{6}} dx = 6 e^{\frac{x}{6}} + C$$ 10. Substitute back: $$\int \frac{x}{3} e^{\frac{x}{6}} dx = 2x e^{\frac{x}{6}} - 2 \times 6 e^{\frac{x}{6}} + C = 2x e^{\frac{x}{6}} - 12 e^{\frac{x}{6}} + C$$ 11. Therefore: $$y e^{\frac{x}{6}} = 2x e^{\frac{x}{6}} - 12 e^{\frac{x}{6}} + C$$ 12. Divide both sides by $$e^{\frac{x}{6}}$$: $$y = 2x - 12 + C e^{-\frac{x}{6}}$$ 13. This is the general solution to the differential equation. Final answer: $$y = 2x - 12 + C e^{-\frac{x}{6}}$$