1. **Problem Statement:** Solve the differential equation $$y' - 2y = \sin(3t)$$ using Laplace transforms. 2. **Formula and Rules:** The Laplace transform of a derivative is $$\mathcal{L}\{y'\} = sY(s) - y(0)$$ where $$Y(s)$$ is the Laplace transform of $$y(t)$$. The Laplace transform of $$\sin(at)$$ is $$\frac{a}{s^2 + a^2}$$. 3. **Apply Laplace transform to both sides:** $$\mathcal{L}\{y' - 2y\} = \mathcal{L}\{\sin(3t)\}$$ $$sY(s) - y(0) - 2Y(s) = \frac{3}{s^2 + 9}$$ 4. **Rearrange to solve for $$Y(s)$$:** $$Y(s)(s - 2) = \frac{3}{s^2 + 9} + y(0)$$ $$Y(s) = \frac{3}{(s - 2)(s^2 + 9)} + \frac{y(0)}{s - 2}$$ 5. **Inverse Laplace transform:** Use partial fraction decomposition and known inverse transforms to find $$y(t)$$. 6. **Final solution:** $$y(t) = y(0)e^{2t} + \int_0^t e^{2(t-\tau)} \sin(3\tau) d\tau$$ which can be evaluated further if initial condition $$y(0)$$ is given. This completes the solution for the first problem.