Integrating both sides with respect to x: y(x) = x³ + 2x + C, where C is the constant of integration.

The characteristic equation is r² - 5r + 6 = 0. Solving for r gives roots r = 2 and r = 3. Thus, the general solution is y(x) = C₁e^(2x) + C₂e^(3x).

Separating variables and integrating, we get y(x) = e^x. Applying the initial condition y(0) = 1, we find C = 1. So, y(x) = e^x.

This is a first-order linear differential equation. The integrating factor is e^x. Multiplying through by e^x and solving, we get y(x) = (1/2)(sin(x) - cos(x)) + Ce^(-x).

The characteristic equation is r² + 4r + 4 = 0, which has a double root r = -2. Thus, the general solution is y(x) = (C₁ + C₂x)e^(-2x).

The general solution to y'' = -λy is y(x) = A sin(√λ x) + B cos(√λ x). Applying boundary conditions, we find B = 0 and √λ = n, where n is an integer. Therefore, y(x) = A sin(nx) for n = 1, 2, 3, ...

Transform the system into matrix form and solve for eigenvalues and eigenvectors. The solution is a combination of exponentials involving the eigenvalues ±5i, leading to solutions of the form x(t) = C₁cos(5t) + C₂sin(5t) and y(t) = C₁sin(5t) - C₂cos(5t).

The particular solution is found using the method of undetermined coefficients. The particular solution is y_p(x) = Ae^(2x), where A is determined by substituting into the original equation, yielding A = 1/5. So, y_p(x) = (1/5)e^(2x).

Solving the homogeneous equation gives y_h(x) = A cos(x) + B sin(x). The solution jumps at x = π due to the delta function, giving a particular solution of y_p(x) = -1/2 sin(x - π) for x > π. Thus, the full solution combines these components.

The Laplace transform of f(t) = e^(-3t) cos(2t) is F(s) = (s + 3)/((s + 3)² + 4).