1. Let's state the problem: Solve a differential equation using the Laplace transform method. 2. The Laplace transform of a function $f(t)$ is defined as $$\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) \, dt.$$ This transform converts differential equations in the time domain into algebraic equations in the $s$-domain. 3. Important rules: - The Laplace transform of a derivative is $$\mathcal{L}\{f'(t)\} = sF(s) - f(0).$$ - For the second derivative, $$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0).$$ 4. Steps to solve: - Take the Laplace transform of both sides of the differential equation. - Substitute initial conditions. - Solve the resulting algebraic equation for $F(s)$. - Use the inverse Laplace transform to find $f(t)$. 5. Example: Solve $$f''(t) + 3f'(t) + 2f(t) = 0$$ with initial conditions $f(0) = 1$, $f'(0) = 0$. 6. Taking Laplace transform: $$s^2F(s) - sf(0) - f'(0) + 3(sF(s) - f(0)) + 2F(s) = 0$$ 7. Substitute initial values: $$s^2F(s) - s(1) - 0 + 3(sF(s) - 1) + 2F(s) = 0$$ 8. Simplify: $$s^2F(s) - s + 3sF(s) - 3 + 2F(s) = 0$$ $$F(s)(s^2 + 3s + 2) = s + 3$$ 9. Solve for $F(s)$: $$F(s) = \frac{s + 3}{s^2 + 3s + 2} = \frac{s + 3}{(s + 1)(s + 2)}$$ 10. Use partial fractions: $$\frac{s + 3}{(s + 1)(s + 2)} = \frac{A}{s + 1} + \frac{B}{s + 2}$$ 11. Multiply both sides by denominator: $$s + 3 = A(s + 2) + B(s + 1)$$ 12. Set $s = -1$: $$-1 + 3 = A(1) + B(0) \Rightarrow 2 = A$$ 13. Set $s = -2$: $$-2 + 3 = A(0) + B(-1) \Rightarrow 1 = -B \Rightarrow B = -1$$ 14. So, $$F(s) = \frac{2}{s + 1} - \frac{1}{s + 2}$$ 15. Take inverse Laplace transform: $$f(t) = 2e^{-t} - e^{-2t}$$ Final answer: $$f(t) = 2e^{-t} - e^{-2t}$$