1. **State the problem:** Solve the initial-value problem $$y' = 3e^{3x - y}, \quad y(0) = \ln(4).$$ 2. **Rewrite the differential equation:** We have $$\frac{dy}{dx} = 3e^{3x - y} = 3e^{3x}e^{-y}.$$ 3. **Separate variables:** Multiply both sides by $e^y$ and $dx$ to get $$e^y dy = 3e^{3x} dx.$$ 4. **Integrate both sides:** $$\int e^y dy = \int 3e^{3x} dx.$$ 5. **Compute the integrals:** Left side: $$\int e^y dy = e^y + C_1,$$ Right side: $$\int 3e^{3x} dx = 3 \cdot \frac{e^{3x}}{3} + C_2 = e^{3x} + C_2.$$ 6. **Combine constants:** Let $C = C_2 - C_1$, then $$e^y = e^{3x} + C.$$ 7. **Apply initial condition:** When $x=0$, $y(0) = \ln(4)$, so $$e^{y(0)} = e^{\ln(4)} = 4 = e^{0} + C = 1 + C,$$ which gives $$C = 3.$$ 8. **Write the explicit solution:** $$e^y = e^{3x} + 3,$$ so $$y = \ln\left(e^{3x} + 3\right).$$ **Final answer:** $$\boxed{y = \ln\left(e^{3x} + 3\right)}.$$