1. **State the problem:** Solve the differential equation $$\frac{dy}{dx} = 2x^2 + y - x^2 y + x y - 2x - 2$$ using the separable variable method. 2. **Rewrite the equation:** Group terms to isolate $y$ and $x$ terms: $$\frac{dy}{dx} = 2x^2 - 2x + y - x^2 y + x y - 2$$ 3. **Factor terms involving $y$:** $$\frac{dy}{dx} = 2x^2 - 2x - 2 + y(1 - x^2 + x)$$ 4. **Rewrite as:** $$\frac{dy}{dx} - y(1 - x^2 + x) = 2x^2 - 2x - 2$$ 5. **Recognize this as a linear first-order ODE:** $$\frac{dy}{dx} + P(x) y = Q(x)$$ where $$P(x) = -(1 - x^2 + x) = x^2 - x - 1$$ and $$Q(x) = 2x^2 - 2x - 2$$ 6. **Find the integrating factor (IF):** $$\mu(x) = e^{\int P(x) dx} = e^{\int (x^2 - x - 1) dx} = e^{\frac{x^3}{3} - \frac{x^2}{2} - x}$$ 7. **Multiply both sides by IF:** $$e^{\frac{x^3}{3} - \frac{x^2}{2} - x} \frac{dy}{dx} + e^{\frac{x^3}{3} - \frac{x^2}{2} - x} (x^2 - x - 1) y = e^{\frac{x^3}{3} - \frac{x^2}{2} - x} (2x^2 - 2x - 2)$$ 8. **Left side is derivative of:** $$\frac{d}{dx} \left(y e^{\frac{x^3}{3} - \frac{x^2}{2} - x}\right) = e^{\frac{x^3}{3} - \frac{x^2}{2} - x} (2x^2 - 2x - 2)$$ 9. **Integrate both sides:** $$y e^{\frac{x^3}{3} - \frac{x^2}{2} - x} = \int e^{\frac{x^3}{3} - \frac{x^2}{2} - x} (2x^2 - 2x - 2) dx + C$$ 10. **Final implicit solution:** $$y = e^{-\frac{x^3}{3} + \frac{x^2}{2} + x} \left( \int e^{\frac{x^3}{3} - \frac{x^2}{2} - x} (2x^2 - 2x - 2) dx + C \right)$$ This integral may not have a simple closed form, but this expression represents the general solution. **Answer:** The solution to the differential equation is given implicitly by the formula above.