1. **State the problem:** Solve the initial value problem $$y''(t) + 4y(t) = t + 4$$ with initial conditions $$y(0) = 1$$ and $$y'(0) = 0$$ using the Laplace transform method. 2. **Recall the Laplace transform properties:** - $$\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)$$ - $$\mathcal{L}\{y(t)\} = Y(s)$$ - $$\mathcal{L}\{t\} = \frac{1}{s^2}$$ - $$\mathcal{L}\{1\} = \frac{1}{s}$$ 3. **Apply Laplace transform to both sides:** $$\mathcal{L}\{y''(t)\} + 4\mathcal{L}\{y(t)\} = \mathcal{L}\{t\} + 4\mathcal{L}\{1\}$$ Substitute: $$s^2Y(s) - s\cdot 1 - 0 + 4Y(s) = \frac{1}{s^2} + \frac{4}{s}$$ 4. **Simplify and solve for $$Y(s)$$:** $$s^2Y(s) - s + 4Y(s) = \frac{1}{s^2} + \frac{4}{s}$$ $$Y(s)(s^2 + 4) = s + \frac{1}{s^2} + \frac{4}{s}$$ Rewrite right side with common denominator $$s^2$$: $$s + \frac{1}{s^2} + \frac{4}{s} = \frac{s^3}{s^2} + \frac{1}{s^2} + \frac{4s}{s^2} = \frac{s^3 + 4s + 1}{s^2}$$ So, $$Y(s) = \frac{s^3 + 4s + 1}{s^2(s^2 + 4)}$$ 5. **Partial fraction decomposition:** Set $$\frac{s^3 + 4s + 1}{s^2(s^2 + 4)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 4}$$ Multiply both sides by $$s^2(s^2 + 4)$$: $$s^3 + 4s + 1 = A s (s^2 + 4) + B (s^2 + 4) + (Cs + D) s^2$$ Expand: $$s^3 + 4s + 1 = A s^3 + 4 A s + B s^2 + 4 B + C s^3 + D s^2$$ Group terms: $$s^3 + 4s + 1 = (A + C) s^3 + (B + D) s^2 + 4 A s + 4 B$$ Equate coefficients: - For $$s^3$$: $$1 = A + C$$ - For $$s^2$$: $$0 = B + D$$ - For $$s^1$$: $$4 = 4 A$$ so $$A = 1$$ - For constant: $$1 = 4 B$$ so $$B = \frac{1}{4}$$ From $$A = 1$$ and $$1 = A + C$$, we get $$C = 0$$. From $$0 = B + D$$, $$D = -B = -\frac{1}{4}$$. 6. **Rewrite $$Y(s)$$:** $$Y(s) = \frac{1}{s} + \frac{1/4}{s^2} + \frac{0 \cdot s - 1/4}{s^2 + 4} = \frac{1}{s} + \frac{1}{4 s^2} - \frac{1/4}{s^2 + 4}$$ 7. **Inverse Laplace transform:** - $$\mathcal{L}^{-1}\{\frac{1}{s}\} = 1$$ - $$\mathcal{L}^{-1}\{\frac{1}{s^2}\} = t$$ - $$\mathcal{L}^{-1}\{\frac{1}{s^2 + 4}\} = \frac{1}{2} \sin(2t)$$ So, $$y(t) = 1 + \frac{1}{4} t - \frac{1}{4} \cdot \frac{1}{2} \sin(2t) = 1 + \frac{t}{4} - \frac{1}{8} \sin(2t)$$ **Final answer:** $$\boxed{y(t) = 1 + \frac{t}{4} - \frac{1}{8} \sin(2t)}$$