1. **State the problem:** Solve the differential equation $$y'' + 2y' + y = xe^{-x}$$ with initial conditions $$y(0) = 1$$ and $$y'(0) = -2$$ using the Laplace transform. 2. **Apply the Laplace transform to both sides:** Recall that $$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$$ and $$\mathcal{L}\{y'\} = sY(s) - y(0)$$ where $$Y(s) = \mathcal{L}\{y(t)\}$$. Applying transforms: $$s^2Y(s) - s\cdot1 - (-2) + 2(sY(s) - 1) + Y(s) = \mathcal{L}\{xe^{-x}\}$$ 3. **Simplify the left side:** $$s^2Y(s) - s + 2 + 2sY(s) - 2 + Y(s) = \mathcal{L}\{xe^{-x}\}$$ Combine constants: $$s^2Y(s) + 2sY(s) + Y(s) - s = \mathcal{L}\{xe^{-x}\}$$ 4. **Factor $$Y(s)$$ terms:** $$Y(s)(s^2 + 2s + 1) - s = \mathcal{L}\{xe^{-x}\}$$ Note that $$s^2 + 2s + 1 = (s+1)^2$$. 5. **Find $$\mathcal{L}\{xe^{-x}\}$$:** Recall $$\mathcal{L}\{t e^{at}\} = \frac{1}{(s - a)^2}$$ for $$s > a$$. Here, $$a = -1$$, so $$\mathcal{L}\{xe^{-x}\} = \frac{1}{(s + 1)^2}$$. 6. **Substitute back:** $$Y(s)(s+1)^2 - s = \frac{1}{(s+1)^2}$$ 7. **Solve for $$Y(s)$$:** $$Y(s)(s+1)^2 = s + \frac{1}{(s+1)^2}$$ Divide both sides by $$(s+1)^2$$: $$Y(s) = \frac{s}{(s+1)^2} + \frac{1}{(s+1)^4}$$ Intermediate step showing cancellation: $$Y(s) = \frac{\cancel{s}}{\cancel{(s+1)^2}} + \frac{1}{(s+1)^4}$$ (no cancellation here, just showing division) 8. **Decompose $$Y(s)$$ for inverse Laplace:** $$Y(s) = \frac{s+1 - 1}{(s+1)^2} + \frac{1}{(s+1)^4} = \frac{s+1}{(s+1)^2} - \frac{1}{(s+1)^2} + \frac{1}{(s+1)^4}$$ Simplify: $$Y(s) = \frac{1}{s+1} - \frac{1}{(s+1)^2} + \frac{1}{(s+1)^4}$$ 9. **Find inverse Laplace transforms:** - $$\mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t}$$ - $$\mathcal{L}^{-1}\{\frac{1}{(s+1)^2}\} = t e^{-t}$$ - $$\mathcal{L}^{-1}\{\frac{1}{(s+1)^4}\} = \frac{t^3}{3!} e^{-t} = \frac{t^3}{6} e^{-t}$$ 10. **Write the solution:** $$y(t) = e^{-t} - t e^{-t} + \frac{t^3}{6} e^{-t} = e^{-t} \left(1 - t + \frac{t^3}{6}\right)$$ 11. **Verify initial conditions:** - $$y(0) = e^{0}(1 - 0 + 0) = 1$$ correct. - $$y'(t) = \text{(differentiate)}$$ and check $$y'(0) = -2$$ (omitted here for brevity but holds true). **Final answer:** $$\boxed{y(t) = e^{-t} \left(1 - t + \frac{t^3}{6}\right)}$$