1. **Stating the problem:** Solve the first-order linear differential equation $$y' = 2ty + 3y - 4t - 6$$ for the function $y(t)$. 2. **Rewrite the equation:** Combine like terms on the right-hand side: $$y' = (2t + 3)y - 4t - 6$$ 3. **Identify the form:** This is a linear ODE of the form $$y' + P(t)y = Q(t)$$ where $$P(t) = -(2t + 3)$$ and $$Q(t) = -4t - 6$$ but to match the standard form, rewrite as: $$y' - (2t + 3)y = -4t - 6$$ 4. **Find the integrating factor (IF):** $$\mu(t) = e^{\int - (2t + 3) dt} = e^{-t^{2} - 3t}$$ 5. **Multiply both sides of the ODE by the integrating factor:** $$e^{-t^{2} - 3t} y' - (2t + 3)e^{-t^{2} - 3t} y = (-4t - 6)e^{-t^{2} - 3t}$$ This simplifies the left side to the derivative of the product: $$\frac{d}{dt} \left(e^{-t^{2} - 3t} y \right) = (-4t - 6)e^{-t^{2} - 3t}$$ 6. **Integrate both sides with respect to $t$:** $$e^{-t^{2} - 3t} y = \int (-4t - 6)e^{-t^{2} - 3t} dt + C$$ 7. **Simplify the integral:** Use substitution. Let $$u = t^{2} + 3t$$ then: $$du = (2t + 3) dt$$ Rewrite the integral: $$\int (-4t - 6) e^{-u} dt = \int -2(2t + 3) e^{-u} dt$$ From $$du = (2t + 3) dt$$, get $$dt = \frac{du}{2t + 3}$$ Then integral becomes: $$-2 \int (2t + 3) e^{-u} \frac{du}{2t + 3} = -2 \int e^{-u} du = -2 \int e^{-u} du$$ 8. **Evaluate the integral:** $$-2 \int e^{-u} du = -2 (-e^{-u}) + C_1 = 2 e^{-u} + C_1$$ 9. **Substitute back $u = t^{2} + 3t$:** $$e^{-t^{2} - 3t} y = 2 e^{-t^{2} - 3t} + C$$ 10. **Multiply both sides by $e^{t^{2} + 3t}$ to solve for $y$:** $$y = 2 + C e^{t^{2} + 3t}$$ **Final answer:** $$\boxed{y(t) = 2 + C e^{t^{2} + 3t}}$$ where $C$ is an arbitrary constant determined by initial conditions if given.