1. **Problem:** Solve the first order linear differential equation $$\frac{dy}{dt} + 4y = 12$$ with initial condition $$y(0) = 2$$. 2. **Formula and rules:** The general form of a first order linear ODE is $$\frac{dy}{dt} + P(t)y = Q(t)$$. The integrating factor (IF) is $$\mu(t) = e^{\int P(t) dt}$$. The solution is given by $$y(t) = \frac{1}{\mu(t)} \left( \int \mu(t) Q(t) dt + C \right)$$. 3. **Identify terms:** Here, $$P(t) = 4$$ and $$Q(t) = 12$$. 4. **Calculate integrating factor:** $$\mu(t) = e^{\int 4 dt} = e^{4t}$$. 5. **Multiply both sides by integrating factor:** $$e^{4t} \frac{dy}{dt} + 4 e^{4t} y = 12 e^{4t}$$ This simplifies to: $$\frac{d}{dt} \left( e^{4t} y \right) = 12 e^{4t}$$ 6. **Integrate both sides:** $$\int \frac{d}{dt} \left( e^{4t} y \right) dt = \int 12 e^{4t} dt$$ $$e^{4t} y = 12 \int e^{4t} dt + C$$ $$e^{4t} y = 12 \cdot \frac{e^{4t}}{4} + C = 3 e^{4t} + C$$ 7. **Solve for $$y$$:** $$y = \frac{3 e^{4t} + C}{e^{4t}} = 3 + C e^{-4t}$$ 8. **Apply initial condition $$y(0) = 2$$:** $$2 = 3 + C e^{0} = 3 + C$$ $$C = 2 - 3 = -1$$ 9. **Final solution:** $$y(t) = 3 - e^{-4t}$$ 10. **Complementary solution $$y_c$$:** Solve homogeneous equation $$\frac{dy}{dt} + 4y = 0$$. $$y_c = C e^{-4t}$$ 11. **Particular solution $$y_p$$:** Constant solution where $$\frac{dy}{dt} = 0$$. $$4 y_p = 12 \Rightarrow y_p = 3$$ **Summary:** - $$y_c = C e^{-4t}$$ - $$y_p = 3$$ - General solution: $$y = y_c + y_p = C e^{-4t} + 3$$ - Definite solution: $$y = 3 - e^{-4t}$$