1. **Problem statement:** Solve the differential equation $$y'' - y' - 6y = 0$$ with initial conditions $$y(0) = 1$$ and $$y'(0) = -1$$ using Laplace transform. 2. **Formula and rules:** The Laplace transform of derivatives are: $$\mathcal{L}\{y'(t)\} = sY(s) - y(0)$$ $$\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)$$ where $$Y(s) = \mathcal{L}\{y(t)\}$$. 3. **Apply Laplace transform to the equation:** $$\mathcal{L}\{y''\} - \mathcal{L}\{y'\} - 6\mathcal{L}\{y\} = 0$$ Substitute transforms: $$s^2Y(s) - sy(0) - y'(0) - (sY(s) - y(0)) - 6Y(s) = 0$$ 4. **Plug in initial values:** $$s^2Y(s) - s(1) - (-1) - sY(s) + 1 - 6Y(s) = 0$$ Simplify: $$s^2Y(s) - s + 1 - sY(s) + 1 - 6Y(s) = 0$$ $$Y(s)(s^2 - s - 6) + (-s + 2) = 0$$ 5. **Solve for $$Y(s)$$:** $$Y(s) = \frac{s - 2}{s^2 - s - 6}$$ Factor denominator: $$s^2 - s - 6 = (s - 3)(s + 2)$$ 6. **Partial fraction decomposition:** $$\frac{s - 2}{(s - 3)(s + 2)} = \frac{A}{s - 3} + \frac{B}{s + 2}$$ Multiply both sides by denominator: $$s - 2 = A(s + 2) + B(s - 3)$$ Set $$s=3$$: $$3 - 2 = A(3 + 2) + B(0) \Rightarrow 1 = 5A \Rightarrow A = \frac{1}{5}$$ Set $$s=-2$$: $$-2 - 2 = A(0) + B(-2 - 3) \Rightarrow -4 = -5B \Rightarrow B = \frac{4}{5}$$ 7. **Rewrite $$Y(s)$$:** $$Y(s) = \frac{1}{5} \cdot \frac{1}{s - 3} + \frac{4}{5} \cdot \frac{1}{s + 2}$$ 8. **Inverse Laplace transform:** $$y(t) = \frac{1}{5}e^{3t} + \frac{4}{5}e^{-2t}$$ **Final answer:** $$\boxed{y(t) = \frac{1}{5}e^{3t} + \frac{4}{5}e^{-2t}}$$