1. **State the problem:** Solve the differential equation $$y'' - 6y' + 9y = 0$$ using the Laplace transform. 2. **Recall the Laplace transform properties:** - The Laplace transform of the derivative is $$\mathcal{L}\{y'\} = sY(s) - y(0)$$. - The Laplace transform of the second derivative is $$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$$. 3. **Apply the Laplace transform to the equation:** $$\mathcal{L}\{y'' - 6y' + 9y\} = \mathcal{L}\{0\}$$ which gives $$s^2Y(s) - sy(0) - y'(0) - 6(sY(s) - y(0)) + 9Y(s) = 0$$ 4. **Rearrange terms:** $$s^2Y(s) - sy(0) - y'(0) - 6sY(s) + 6y(0) + 9Y(s) = 0$$ 5. **Group terms with $$Y(s)$$:** $$Y(s)(s^2 - 6s + 9) = sy(0) + y'(0) - 6y(0)$$ 6. **Simplify the characteristic polynomial:** $$s^2 - 6s + 9 = (s - 3)^2$$ 7. **Solve for $$Y(s)$$:** $$Y(s) = \frac{sy(0) + y'(0) - 6y(0)}{(s - 3)^2}$$ 8. **Assuming initial conditions $$y(0) = y_0$$ and $$y'(0) = y_1$$, write:** $$Y(s) = \frac{s y_0 + y_1 - 6 y_0}{(s - 3)^2} = \frac{y_0 (s - 6) + y_1}{(s - 3)^2}$$ 9. **Use inverse Laplace transform formulas:** - $$\mathcal{L}^{-1}\{\frac{1}{(s - a)^2}\} = t e^{at}$$ - $$\mathcal{L}^{-1}\{\frac{s - a}{(s - a)^2}\} = e^{at}$$ 10. **Rewrite numerator:** $$y_0 (s - 6) + y_1 = y_0 (s - 3 - 3) + y_1 = y_0 (s - 3) - 3 y_0 + y_1$$ 11. **Split $$Y(s)$$:** $$Y(s) = y_0 \frac{s - 3}{(s - 3)^2} + \frac{y_1 - 3 y_0}{(s - 3)^2}$$ 12. **Apply inverse Laplace transform:** $$y(t) = y_0 e^{3t} + (y_1 - 3 y_0) t e^{3t}$$ **Final answer:** $$\boxed{y(t) = y_0 e^{3t} + (y_1 - 3 y_0) t e^{3t}}$$ This is the general solution to the differential equation using Laplace transforms.