1. **State the problem:** Solve the second-order linear differential equation $$y'' - 2y' - 8y = 0$$ with initial conditions $$y(0) = 3$$ and $$y'(0) = 6$$. 2. **Characteristic equation:** For a differential equation of the form $$y'' + ay' + by = 0$$, the characteristic equation is $$r^2 + ar + b = 0$$. Here, $$a = -2$$ and $$b = -8$$, so the characteristic equation is: $$r^2 - 2r - 8 = 0$$. 3. **Solve the characteristic equation:** Use the quadratic formula: $$r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2}$$ $$r = \frac{2 \pm 6}{2}$$ So the roots are: $$r_1 = \frac{2 + 6}{2} = 4$$ $$r_2 = \frac{2 - 6}{2} = -2$$ 4. **General solution:** Since roots are real and distinct, the general solution is: $$y(t) = C_1 e^{4t} + C_2 e^{-2t}$$ 5. **Apply initial conditions:** - At $$t=0$$, $$y(0) = C_1 + C_2 = 3$$ - Derivative: $$y'(t) = 4C_1 e^{4t} - 2C_2 e^{-2t}$$ At $$t=0$$, $$y'(0) = 4C_1 - 2C_2 = 6$$ 6. **Solve system of equations:** From $$C_1 + C_2 = 3$$, we get $$C_2 = 3 - C_1$$. Substitute into $$4C_1 - 2C_2 = 6$$: $$4C_1 - 2(3 - C_1) = 6$$ $$4C_1 - 6 + 2C_1 = 6$$ $$6C_1 = 12$$ $$C_1 = 2$$ Then, $$C_2 = 3 - 2 = 1$$ 7. **Final solution:** $$\boxed{y(t) = 2 e^{4t} + e^{-2t}}$$