1. **State the problem:** Solve the second order ODE $$\frac{d^2y}{dt^2} - 6\frac{dy}{dt} + 16y = 2e^{2t}$$ with initial conditions $$y(0) = -1$$ and $$y'(0) = -4$$ using Laplace transform. 2. **Recall Laplace transform properties:** - $$\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)$$ - $$\mathcal{L}\{y'(t)\} = sY(s) - y(0)$$ - $$\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$$ 3. **Apply Laplace transform to both sides:** $$\mathcal{L}\{y'' - 6y' + 16y\} = \mathcal{L}\{2e^{2t}\}$$ $$s^2Y(s) - s(-1) - (-4) - 6(sY(s) - (-1)) + 16Y(s) = \frac{2}{s-2}$$ 4. **Simplify the left side:** $$s^2Y(s) + s + 4 - 6sY(s) - 6 + 16Y(s) = \frac{2}{s-2}$$ $$Y(s)(s^2 - 6s + 16) + (s + 4 - 6) = \frac{2}{s-2}$$ $$Y(s)(s^2 - 6s + 16) + (s - 2) = \frac{2}{s-2}$$ 5. **Isolate $$Y(s)$$:** $$Y(s)(s^2 - 6s + 16) = \frac{2}{s-2} - (s - 2)$$ 6. **Combine right side:** $$\frac{2}{s-2} - (s - 2) = \frac{2 - (s-2)(s-2)}{s-2} = \frac{2 - (s-2)^2}{s-2}$$ 7. **Expand $$ (s-2)^2 $$:** $$ (s-2)^2 = s^2 - 4s + 4 $$ So numerator: $$ 2 - (s^2 - 4s + 4) = 2 - s^2 + 4s - 4 = -s^2 + 4s - 2 $$ 8. **Rewrite $$Y(s)$$:** $$Y(s) = \frac{-s^2 + 4s - 2}{(s-2)(s^2 - 6s + 16)}$$ 9. **Factor denominator quadratic:** $$s^2 - 6s + 16$$ has discriminant $$\Delta = (-6)^2 - 4 \times 1 \times 16 = 36 - 64 = -28 < 0$$ so no real roots. 10. **Partial fraction decomposition:** Set $$\frac{-s^2 + 4s - 2}{(s-2)(s^2 - 6s + 16)} = \frac{A}{s-2} + \frac{Bs + C}{s^2 - 6s + 16}$$ Multiply both sides by denominator: $$-s^2 + 4s - 2 = A(s^2 - 6s + 16) + (Bs + C)(s - 2)$$ 11. **Expand right side:** $$A s^2 - 6A s + 16 A + B s^2 - 2 B s + C s - 2 C$$ Group terms: $$ (A + B) s^2 + (-6A - 2B + C) s + (16 A - 2 C)$$ 12. **Equate coefficients:** - For $$s^2$$: $$A + B = -1$$ - For $$s$$: $$-6A - 2B + C = 4$$ - Constant: $$16 A - 2 C = -2$$ 13. **Solve system:** From first: $$B = -1 - A$$ Substitute into second: $$-6A - 2(-1 - A) + C = 4 \Rightarrow -6A + 2 + 2A + C = 4 \Rightarrow -4A + C = 2$$ From third: $$16 A - 2 C = -2 \Rightarrow -2 C = -2 - 16 A \Rightarrow C = 8 A + 1$$ Substitute $$C$$ into second: $$-4 A + 8 A + 1 = 2 \Rightarrow 4 A + 1 = 2 \Rightarrow 4 A = 1 \Rightarrow A = \frac{1}{4}$$ Then $$B = -1 - \frac{1}{4} = -\frac{5}{4}$$ $$C = 8 \times \frac{1}{4} + 1 = 2 + 1 = 3$$ 14. **Rewrite $$Y(s)$$:** $$Y(s) = \frac{1/4}{s-2} + \frac{-\frac{5}{4} s + 3}{s^2 - 6 s + 16}$$ 15. **Complete the square in denominator:** $$s^2 - 6 s + 16 = (s - 3)^2 + 7$$ 16. **Rewrite numerator of second term:** $$-\frac{5}{4} s + 3 = -\frac{5}{4} (s - 3) + \left(3 - \frac{5}{4} \times (-3)\right) = -\frac{5}{4} (s - 3) + \left(3 + \frac{15}{4}\right) = -\frac{5}{4} (s - 3) + \frac{27}{4}$$ 17. **Split second term:** $$\frac{-\frac{5}{4} s + 3}{(s - 3)^2 + 7} = -\frac{5}{4} \frac{s - 3}{(s - 3)^2 + 7} + \frac{27}{4} \frac{1}{(s - 3)^2 + 7}$$ 18. **Recall inverse Laplace transforms:** - $$\mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{a t}$$ - $$\mathcal{L}^{-1}\left\{\frac{s - a}{(s - a)^2 + b^2}\right\} = e^{a t} \cos(b t)$$ - $$\mathcal{L}^{-1}\left\{\frac{b}{(s - a)^2 + b^2}\right\} = e^{a t} \sin(b t)$$ 19. **Apply inverse Laplace:** $$y(t) = \frac{1}{4} e^{2 t} - \frac{5}{4} e^{3 t} \cos(\sqrt{7} t) + \frac{27}{4} \times \frac{1}{\sqrt{7}} e^{3 t} \sin(\sqrt{7} t)$$ 20. **Final answer:** $$\boxed{y(t) = \frac{1}{4} e^{2 t} - \frac{5}{4} e^{3 t} \cos(\sqrt{7} t) + \frac{27}{4 \sqrt{7}} e^{3 t} \sin(\sqrt{7} t)}$$