1. The problem is to solve the differential equation $$x'' + 4x' + 5x = 80\sin 5t$$ with initial conditions $$x(0) = 0$$ and $$x'(0) = 0$$. 2. This is a second-order linear nonhomogeneous differential equation with constant coefficients. The general solution is $$x(t) = x_h(t) + x_p(t)$$ where $$x_h$$ is the homogeneous solution and $$x_p$$ is a particular solution. 3. First, solve the homogeneous equation $$x'' + 4x' + 5x = 0$$. 4. The characteristic equation is $$r^2 + 4r + 5 = 0$$. 5. Solve for $$r$$ using the quadratic formula: $$r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i$$. 6. So the homogeneous solution is: $$x_h(t) = e^{-2t}(C_1 \cos t + C_2 \sin t)$$. 7. Next, find a particular solution $$x_p(t)$$ for the right side $$80 \sin 5t$$. 8. Since the forcing function is $$80 \sin 5t$$, try a particular solution of the form: $$x_p(t) = A \cos 5t + B \sin 5t$$. 9. Compute derivatives: $$x_p' = -5A \sin 5t + 5B \cos 5t$$ $$x_p'' = -25A \cos 5t - 25B \sin 5t$$. 10. Substitute into the differential equation: $$x_p'' + 4x_p' + 5x_p = (-25A \cos 5t - 25B \sin 5t) + 4(-5A \sin 5t + 5B \cos 5t) + 5(A \cos 5t + B \sin 5t)$$ 11. Group terms: $$= (-25A + 20B + 5A) \cos 5t + (-25B - 20A + 5B) \sin 5t = (-20A + 20B) \cos 5t + (-20B - 20A) \sin 5t$$. 12. Set equal to the right side $$80 \sin 5t$$: $$(-20A + 20B) \cos 5t + (-20B - 20A) \sin 5t = 0 \cdot \cos 5t + 80 \sin 5t$$. 13. Equate coefficients: $$-20A + 20B = 0$$ $$-20B - 20A = 80$$. 14. Solve the system: From first: $$-20A + 20B = 0 \Rightarrow B = A$$. Substitute into second: $$-20A - 20A = 80 \Rightarrow -40A = 80 \Rightarrow A = -2$$. Then $$B = -2$$. 15. So the particular solution is: $$x_p(t) = -2 \cos 5t - 2 \sin 5t$$. 16. The general solution is: $$x(t) = e^{-2t}(C_1 \cos t + C_2 \sin t) - 2 \cos 5t - 2 \sin 5t$$. 17. Apply initial conditions: $$x(0) = C_1 - 2 = 0 \Rightarrow C_1 = 2$$. 18. Compute $$x'(t)$$: $$x'(t) = \frac{d}{dt} \left[e^{-2t}(2 \cos t + C_2 \sin t) - 2 \cos 5t - 2 \sin 5t\right]$$ 19. Differentiate: $$x'(t) = e^{-2t}(-2)(2 \cos t + C_2 \sin t) + e^{-2t}(-2 \sin t + C_2 \cos t) + 10 \sin 5t - 10 \cos 5t$$ 20. Simplify: $$x'(t) = e^{-2t}[-4 \cos t - 2 C_2 \sin t - 2 \sin t + C_2 \cos t] + 10 \sin 5t - 10 \cos 5t$$ 21. Evaluate at $$t=0$$: $$x'(0) = [-4 + C_2] + 0 - 10 = -4 + C_2 - 10 = C_2 - 14$$ 22. Given $$x'(0) = 0$$, solve: $$C_2 - 14 = 0 \Rightarrow C_2 = 14$$. 23. Final solution: $$\boxed{x(t) = e^{-2t}(2 \cos t + 14 \sin t) - 2 \cos 5t - 2 \sin 5t}$$