A. The characteristic equation for the associated homogeneous equation can be found by setting the coefficients of the derivatives to zero. In this case, the equation is y'' - 4y' + 3y = 0. To find the characteristic equation, we replace the derivatives with r (our variable) to get r^2 - 4r + 3 = 0.
B. To find the fundamental solutions for the associated homogeneous equation, we solve the characteristic equation. Factoring the equation, we have (r - 1)(r - 3) = 0. Therefore, the fundamental solutions are y1 = e^t and y2 = e^(3t).
C. To find the form of the particular solution and its derivatives, we need to consider the form of the nonhomogeneous term, 12e^t - 12te^t - (6t + 1). As this term contains e^t, e^(3t), and t, we assume a particular solution of the form Y = Ae^t + Be^(3t) + Ct + D, where A, B, C, and D are undetermined coefficients. Taking the first and second derivatives of Y, we get Y' = Ae^t + 3Be^(3t) + C and Y'' = Ae^t + 9Be^(3t).
D. The general solution can be obtained by combining the fundamental solutions and the particular solution. Therefore, the general solution is y = c1e^t + c2e^(3t) + Ae^t + Be^(3t) + Ct + D, where c1 and c2 are constants.
E. To find the values of c1 and c2 and solve the initial value problem, we substitute the initial values y(0) = 1 and y'(0) = 1 into the general solution. This gives us the equations 1 = c1 + c2 + A + D and 1 = c1 + 3c2 + A. Solving these equations will give us the values of c1 and c2, which can be used to find the solution to the initial value problem.