1. **State the problem:** Solve the initial value problem using the Laplace transform: $$y'' + 3y' = 0, \quad y(0) = 6, \quad y'(0) = 4$$ 2. **Apply the Laplace transform:** Let $Y(s) = \mathcal{L}\{y(t)\}$. Using linearity and the transforms of derivatives: $$\mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0) = s^2 Y(s) - 6s - 4$$ $$\mathcal{L}\{y'\} = s Y(s) - y(0) = s Y(s) - 6$$ 3. **Transform the differential equation:** $$s^2 Y(s) - 6s - 4 + 3(s Y(s) - 6) = 0$$ 4. **Simplify the equation:** $$s^2 Y(s) - 6s - 4 + 3s Y(s) - 18 = 0$$ Group terms with $Y(s)$ and constants: $$(s^2 + 3s) Y(s) - 6s - 4 - 18 = 0$$ $$(s^2 + 3s) Y(s) = 6s + 22$$ 5. **Solve for $Y(s)$:** $$Y(s) = \frac{6s + 22}{s^2 + 3s} = \frac{6s + 22}{s(s + 3)}$$ 6. **Partial fraction decomposition:** Assume: $$\frac{6s + 22}{s(s + 3)} = \frac{A}{s} + \frac{B}{s + 3}$$ Multiply both sides by $s(s+3)$: $$6s + 22 = A(s + 3) + Bs$$ Set $s=0$: $$22 = 3A \implies A = \frac{22}{3}$$ Set $s = -3$: $$6(-3) + 22 = B(-3) \implies -18 + 22 = -3B \implies 4 = -3B \implies B = -\frac{4}{3}$$ 7. **Rewrite $Y(s)$:** $$Y(s) = \frac{22/3}{s} - \frac{4/3}{s + 3}$$ 8. **Inverse Laplace transform:** Recall: $$\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1$$ $$\mathcal{L}^{-1}\left\{\frac{1}{s + a}\right\} = e^{-at}$$ So, $$y(t) = \frac{22}{3} - \frac{4}{3} e^{-3t}$$ **Final answer:** $$\boxed{y(t) = \frac{22}{3} - \frac{4}{3} e^{-3t}}$$