1. **Problem:** Solve the ODE $$(x^2 - 4) y' = -2xy - 6x$$ 2. **Rewrite the equation:** $$y' = \frac{-2xy - 6x}{x^2 - 4}$$ 3. **Separate variables or identify the type:** This is a first-order linear ODE in $y$ with respect to $x$. 4. **Rewrite as:** $$y' + \frac{2x}{x^2 - 4} y = \frac{-6x}{x^2 - 4}$$ 5. **Find the integrating factor (IF):** $$\mu(x) = e^{\int \frac{2x}{x^2 - 4} dx}$$ 6. **Calculate the integral in the exponent:** Let $u = x^2 - 4$, then $du = 2x dx$, so $$\int \frac{2x}{x^2 - 4} dx = \int \frac{1}{u} du = \ln|u| = \ln|x^2 - 4|$$ 7. **Integrating factor:** $$\mu(x) = e^{\ln|x^2 - 4|} = |x^2 - 4|$$ 8. **Multiply both sides by $\mu(x)$:** $$|x^2 - 4| y' + \frac{2x}{x^2 - 4} |x^2 - 4| y = -6x$$ Simplifies to: $$(x^2 - 4) y' + 2x y = -6x$$ 9. **Recognize left side as derivative:** $$\frac{d}{dx} \big((x^2 - 4) y\big) = -6x$$ 10. **Integrate both sides:** $$\int \frac{d}{dx} \big((x^2 - 4) y\big) dx = \int -6x dx$$ $$ (x^2 - 4) y = -3x^2 + C$$ 11. **Solve for $y$:** $$y = \frac{-3x^2 + C}{x^2 - 4}$$ **Final answer:** $$\boxed{y = \frac{C - 3x^2}{x^2 - 4}}$$ This completes the solution for the first ODE.