Final Answer:
The particular solution to the given differential equation is y = 6e^4t + 4e^0t.
Step-by-step explanation:
The given differential equation is of the form y'' - 8y' + 34y = g(t), where g(t) = 80e^4tcos(4t) + 48e^4tsin(4t) + 8e^0t. This is a second order linear differential equation with constant coefficients, and can be solved using the method of undetermined coefficients.
For this method, we need to find the solution of the associated homogeneous equation first, which is y'' - 8y' + 34y = 0. The characteristic equation of this equation is r^2 - 8r + 34 = 0, which has two distinct roots at r = 6 and r = 2. Therefore, the general solution of the associated homogeneous equation is yh = c1e^6t + c2e^2t.
Now, to find the particular solution, we need to guess the form of the solution. Since g(t) is a combination of cos(4t) and sin(4t), we assume that the particular solution contains terms of the form Ae^4tcos(4t) + Be^4tsin(4t) + Ce^0t. Substituting this into the differential equation, we get A(16 - 8A)e^4tcos(4t) + B(16 - 8B)e^4tsin(4t) + (C - 8A - 8B)e^0t = g(t).
Equating the coefficients of e^4tcos(4t), e^4tsin(4t) and e^0t on both sides, we get A = 6, B = 4 and C = 8. Therefore, the particular solution of the given differential equation is y = 6e^4tcos(4t) + 4e^4tsin(4t) + 8e^0t. Since cos(4t) and sin(4t) are periodic functions, the constants A and B can be replaced with 6e^4t and 4e^4t, respectively. Thus, the particular solution to the given differential equation is y = 6e^4t + 4e^0t.