1. **Problem statement:** Solve the initial value problem (IVP) by inverting the Laplace transform for the differential equation $$y'' - 3y' - 10y = 1$$ with initial conditions $$y(0) = -1$$ and $$y'(0) = 2$$. 2. **Formula and rules:** Taking the Laplace transform of both sides, recall that $$\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$$ and $$\mathcal{L}\{y'\} = sY(s) - y(0)$$ where $$Y(s) = \mathcal{L}\{y(t)\}$$. 3. **Apply Laplace transform:** $$s^2Y(s) - s(-1) - 2 - 3\left(sY(s) - (-1)\right) - 10Y(s) = \frac{1}{s}$$ 4. **Simplify:** $$s^2Y(s) + s - 2 - 3sY(s) - 3 + -10Y(s) = \frac{1}{s}$$ $$\Rightarrow (s^2 - 3s - 10)Y(s) + (s - 5) = \frac{1}{s}$$ 5. **Isolate $$Y(s)$$:** $$ (s^2 - 3s - 10)Y(s) = \frac{1}{s} - s + 5 $$ 6. **Rewrite right side:** $$Y(s) = \frac{\frac{1}{s} - s + 5}{s^2 - 3s - 10}$$ 7. **Factor denominator:** $$s^2 - 3s - 10 = (s - 5)(s + 2)$$ 8. **Express $$Y(s)$$ as:** $$Y(s) = \frac{1/s - s + 5}{(s - 5)(s + 2)} = \frac{1/s}{(s - 5)(s + 2)} - \frac{s - 5}{(s - 5)(s + 2)} = \frac{1}{s(s - 5)(s + 2)} - \frac{1}{s + 2}$$ 9. **Partial fraction decomposition for $$\frac{1}{s(s - 5)(s + 2)}$$:** Assume $$\frac{1}{s(s - 5)(s + 2)} = \frac{A}{s} + \frac{B}{s - 5} + \frac{C}{s + 2}$$ Multiply both sides by $$s(s - 5)(s + 2)$$: $$1 = A(s - 5)(s + 2) + Bs(s + 2) + Cs(s - 5)$$ 10. **Find coefficients:** - Let $$s=0$$: $$1 = A(-5)(2) = -10A \Rightarrow A = -\frac{1}{10}$$ - Let $$s=5$$: $$1 = B(5)(7) = 35B \Rightarrow B = \frac{1}{35}$$ - Let $$s=-2$$: $$1 = C(-2)(-7) = 14C \Rightarrow C = \frac{1}{14}$$ 11. **Rewrite $$Y(s)$$:** $$Y(s) = -\frac{1}{10}\frac{1}{s} + \frac{1}{35}\frac{1}{s - 5} + \frac{1}{14}\frac{1}{s + 2} - \frac{1}{s + 2}$$ 12. **Combine terms with $$\frac{1}{s + 2}$$:** $$\frac{1}{14} - 1 = -\frac{13}{14}$$ 13. **Final form:** $$Y(s) = -\frac{1}{10}\frac{1}{s} + \frac{1}{35}\frac{1}{s - 5} - \frac{13}{14}\frac{1}{s + 2}$$ 14. **Inverse Laplace transform:** Recall $$\mathcal{L}^{-1}\{\frac{1}{s}\} = 1$$, $$\mathcal{L}^{-1}\{\frac{1}{s - a}\} = e^{at}$$. 15. **Solution:** $$y(t) = -\frac{1}{10} + \frac{1}{35}e^{5t} - \frac{13}{14}e^{-2t}$$ **Final answer:** $$\boxed{y(t) = -\frac{1}{10} + \frac{1}{35}e^{5t} - \frac{13}{14}e^{-2t}}$$