1. **State the problem:** Solve the differential equation $$y'' - 4y = 8e^{3t}$$ with initial conditions $$y(0) = 4$$ and $$y'(0) = 0$$ using Laplace transforms. 2. **Take the Laplace transform of both sides:** Using linearity and the property $$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$$, we get: $$s^2Y(s) - 4Y(s) - 4s = \frac{8}{s-3}$$ 3. **Substitute initial conditions:** $$y(0) = 4, y'(0) = 0$$ so $$s^2Y(s) - 4Y(s) - 4s = \frac{8}{s-3}$$ 4. **Rearrange to solve for $$Y(s)$$:** $$Y(s)(s^2 - 4) = \frac{8}{s-3} + 4s$$ $$Y(s) = \frac{8}{(s-3)(s^2 - 4)} + \frac{4s}{s^2 - 4}$$ 5. **Factor denominator:** $$s^2 - 4 = (s-2)(s+2)$$ 6. **Partial fraction decomposition:** Write $$\frac{8}{(s-3)(s-2)(s+2)} = \frac{A}{s-3} + \frac{B}{s-2} + \frac{C}{s+2}$$ Multiply both sides by denominator: $$8 = A(s-2)(s+2) + B(s-3)(s+2) + C(s-3)(s-2)$$ 7. **Find coefficients:** Set $$s=3$$: $$8 = A(1)(5) = 5A \Rightarrow A = \frac{8}{5}$$ Set $$s=2$$: $$8 = B(-1)(4) = -4B \Rightarrow B = -2$$ Set $$s=-2$$: $$8 = C(-5)(-4) = 20C \Rightarrow C = \frac{2}{5}$$ 8. **Rewrite $$Y(s)$$:** $$Y(s) = \frac{8}{5(s-3)} - \frac{2}{s-2} + \frac{2}{5(s+2)} + \frac{4s}{(s-2)(s+2)}$$ 9. **Decompose $$\frac{4s}{(s-2)(s+2)}$$:** Write $$\frac{4s}{(s-2)(s+2)} = \frac{D}{s-2} + \frac{E}{s+2}$$ Multiply both sides: $$4s = D(s+2) + E(s-2)$$ Set $$s=2$$: $$8 = D(4) \Rightarrow D=2$$ Set $$s=-2$$: $$-8 = E(-4) \Rightarrow E=2$$ 10. **Combine terms:** $$Y(s) = \frac{8}{5(s-3)} - \frac{2}{s-2} + \frac{2}{5(s+2)} + \frac{2}{s-2} + \frac{2}{s+2}$$ Simplify: $$Y(s) = \frac{8}{5(s-3)} + \left(-\frac{2}{s-2} + \frac{2}{s-2}\right) + \left(\frac{2}{5(s+2)} + \frac{2}{s+2}\right)$$ $$Y(s) = \frac{8}{5(s-3)} + 0 + \frac{2}{5(s+2)} + \frac{10}{5(s+2)} = \frac{8}{5(s-3)} + \frac{12}{5(s+2)}$$ 11. **Take inverse Laplace transform:** $$y(t) = \frac{8}{5}e^{3t} + \frac{12}{5}e^{-2t}$$ 12. **Final answer:** $$\boxed{y = \frac{12}{5}e^{-2t} + \frac{8}{5}e^{3t}}$$ This matches option E.