1. **State the problem:** Solve the differential equation $$y'' - 9y = -24e^{-2t}$$ with initial conditions $$y(0) = 8$$ and $$y'(0) = 0$$. 2. **Solve the homogeneous equation:** The associated homogeneous equation is $$y'' - 9y = 0$$. The characteristic equation is $$r^2 - 9 = 0$$, which factors as $$(r-3)(r+3) = 0$$. So, the roots are $$r = 3$$ and $$r = -3$$. The general solution to the homogeneous equation is $$y_h = C_1 e^{3t} + C_2 e^{-3t}$$. 3. **Find a particular solution:** Since the right side is $$-24 e^{-2t}$$, try a particular solution of the form $$y_p = A e^{-2t}$$. Calculate derivatives: $$y_p' = -2A e^{-2t}$$ $$y_p'' = 4A e^{-2t}$$. Substitute into the differential equation: $$4A e^{-2t} - 9 A e^{-2t} = -24 e^{-2t}$$ $$(-5A) e^{-2t} = -24 e^{-2t}$$ Divide both sides by $$e^{-2t}$$ (nonzero): $$-5A = -24$$ Solve for $$A$$: $$A = \frac{24}{5}$$. So, the particular solution is $$y_p = \frac{24}{5} e^{-2t}$$. 4. **Write the general solution:** $$y = y_h + y_p = C_1 e^{3t} + C_2 e^{-3t} + \frac{24}{5} e^{-2t}$$. 5. **Apply initial conditions:** At $$t=0$$: $$y(0) = C_1 + C_2 + \frac{24}{5} = 8$$ So, $$C_1 + C_2 = 8 - \frac{24}{5} = \frac{40}{5} - \frac{24}{5} = \frac{16}{5}$$. 6. **Find $$y'(t)$$:** $$y' = 3 C_1 e^{3t} - 3 C_2 e^{-3t} - \frac{48}{5} e^{-2t}$$. At $$t=0$$: $$y'(0) = 3 C_1 - 3 C_2 - \frac{48}{5} = 0$$ Rearranged: $$3 C_1 - 3 C_2 = \frac{48}{5}$$ Divide both sides by 3: $$C_1 - C_2 = \frac{16}{5}$$. 7. **Solve the system:** $$C_1 + C_2 = \frac{16}{5}$$ $$C_1 - C_2 = \frac{16}{5}$$ Add equations: $$2 C_1 = \frac{32}{5} \Rightarrow C_1 = \frac{16}{5}$$ Substitute back: $$\frac{16}{5} + C_2 = \frac{16}{5} \Rightarrow C_2 = 0$$. 8. **Final solution:** $$y = \frac{16}{5} e^{3t} + 0 \cdot e^{-3t} + \frac{24}{5} e^{-2t} = \frac{16}{5} e^{3t} + \frac{24}{5} e^{-2t}$$. 9. **Match with options:** This corresponds to option B. **Answer:** $$y = \frac{16}{5} e^{3t} + \frac{24}{5} e^{-2t}$$